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UNIVERSITY  OF  CALIFORNIA. 


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MANUAL 


PLANE    GEOMETRY, 


ON  THE  HEUKISTIC  PLAN, 


WITH  NUMEROUS  EXTRA   EXERCISES,  BOTH 

THEOREMS  AND  PROBLEMS,  FOR 

ADVANCE  WORK. 


BY 


G.  IRVING  HOPKINS, 

INSTRUCTOR  IN  MATHEMATICS  AND  PHYSICS  IN  HIGH 
SCHOOL,  MANCHESTER,  N.H. 


BOSTON,  U.S.A.: 

D.   C.   HEATH  &  COMPANY. 

1891. 


COPYRIGHT,  1891, 
BY  G.  I.  HOPKINS. 


TYPOGRAPHY  BY  J.  S.  GUSHING  &  Co.,  BOSTON,  U.S.A. 

PRESSWORK  BY  BERWICK  &  SMITH,   BOSTON,  U.S.A. 


PREFACE. 


THIS  book  is  published  primarily  for  the  author's  pupils, 
and  secondarily  for  that  constantly  increasing  number  of 
teachers  who  are  getting  more  and  more  dissatisfied  with 
the  old  methods  of  teaching  geometry,  but  who  have  hitherto 
found  no  manual  suited  to  their  needs.  That  the  reasoning 
faculties  of  a  child  or  youth  are  developed  by  use,  goes 
without  saying;  and  consequently  the  method  of  teaching 
geometry  whereby  the  pupil  originates  the  demonstration,  or 
simply  demonstrates,  rather  than  memorizes  the  demonstrations 
of  another,  needs  no  defence  at  my  hands.  This  manual  has 
stood  the  test  of  three  years'  work  in  the  class-room,  and 
differs  from  other  geometries  that  provide  for  original  work, 
in  that  the  original  demonstrations  and  constructions  are 
not  side  issues,  so  to  speak,  but  are  required  of  the  pupil  in 
that  sequence  of  theorems  and  problems  which  may  be  called 
the  regular  required  work,  and  of  which  complete  demonstra- 
tions are  usually  given  in  other  manuals  for  the  pupil  to 
memorize.  In  this  work  demonstrations  are  given  only  where 
the  average  pupil  would  be  at  a  loss  to  know  how  to  proceed, 
and  generally,  as  illustrative  of  methods,  while  others  are 
partially  given  and  left  for  the  pupil  to  complete.  In  other 
cases,  wherever  the  author  has  found  that  his  own  pupils 
were  not  working  to  advantage,  he  has  introduced  sugges- 
tions, of  which  pupils  may  or  may  not  avail  themselves. 

iii 

183668 


iv  PREFACE. 

The  old  method  of  division  into  books  has  been  abandoned, 
as  it  serves  no  practical  use,  since  different  authors  have  made 
different  divisions.  It  serves  to  make  the  subject  a  continu- 
ous one  in  the  mind  of  the  pupil,  without  the  artificial  breaks 
of  the  old  way. 

The  arrangement  of  the  theorems  whereby  the  essential 
ones,  together  with  several  simple  additional  ones  for  giving 
the  pupil  more  drill,  followed  by  a  set  of  non-essential  but 
more  difficult  ones  for  advance  work,  the  author  has  found 
advantageous. 

He  has  also  found  it  more  useful  to  the  pupil  to  be  com- 
pelled to  construct  his  own  diagram,  and  state  the  converse 
of  the  theorems  he  is  to  prove.  Practical  problems  of  com- 
putation have  also  been  given,  immediately  following  any 
given  subject,  so  that  the  pupil  can  immediately  see  the 
practical  application  of  the  theorems  he  has  demonstrated. 

All  the  problems  of  construction  have  been  placed  together 
after  the  theorems,  for  the  sake  of  uniformity,  as  they  form 
no  part  of  the  logical  sequence  of  geometrical  truths  as  em- 
bodied in  the  theorems.  Many  of  them  are  of  practical,  as 
well  as  disciplinary,  importance,  however,  and  so  it  is  left 
to  the  skilful  and  judicious  teacher  to  take  them  up  as  the 
interests  of  his  pupils  demand. 

The  author  has  received  valuable  suggestions  from  Pro- 
fessor E.  L.  Richards  of  Yale  College,  and  also  from  Profes- 
sor T.  H.  Safford  of  Williams  College.  He  also  desires  to 
publicly  express  his  indebtedness  to  Hon.  J.  W.  Patterson, 
Superintendent  of  Public  Instruction  for  New  Hampshire, 
and  E.  R.  Goodwin,  Principal  of  the  High  School  at  Lawrence, 
Mass.,  for  their  hearty  and  outspoken  indorsement  of  his 


PEEFACE.  V 

work,  and  encouragement  to  persevere  in  elaborating  the 
method. 

Finally,  thanks  are  due  the  publishers  and  printers  for 
the  excellence  and  beauty  of  the  mechanical  work. 

In  conclusion,  the  author  would  be  glad  to  receive  sugges- 
tions from  those  teachers  into  whose  hands  this  manual  may 
chance  to  fall,  with  a  view  to  its  improvement  as  a  regular 
class  text-book. 


G.  I.  H. 


MANCHESTER,  N.H. 
May,  1891. 


TABLE   OF   CONTENTS. 


PAGE 

DEFINITIONS 1 

GENERAL  AXIOMS ~~& 

PARTICULAR  AXIOMS — 7 

SYMBOLS 10 

ABBREVIATIONS 11 

THEOREMS 12 

ANGLE  MEASUREMENT 15 

TRANSVERSALS 20 

TRIANGLES 24 

ADVANCE  THEOREMS 32,  35,  52,  80,  94, 117 

QUADRILATERALS 33 

CIRCLES 38 

RATIO  AND  PROPORTION 54 

PROPORTIONAL  LINES 68 

POLYGONS 71 

PROBLEMS  OF  COMPUTATION 72, 82,  95, 119 

SIMILAR  FIGURES 74 

SIMILAR  TRIANGLES 74 

PROJECTION 78 

AREAS 84 

REGULAR  POLYGONS  AND  CIRCLES 98 

MAXIMA  AND  MINIMA 124 

PROBLEMS  OF  CONSTRUCTION 133 

PLANE  PROBLEMS 133 

OPTIONAL  PROBLEMS  FOR  ADVANCE  WORK 143 

REQUIRED  PROBLEMS  OF  CONSTRUCTION 147 

vii 


Vlil  TABLE  OF  CONTENTS. 

MISCELLANEOUS  PLANE  PROBLEMS  FOB  ADVANCE  WORK:  PAQB 

I.  Triangles 155 

II.  Quadrilaterals 158 

III.  Circles 159 

IV.  Transformation  of  Figures 160 

V.  Division  of  Figures 163 

APPENDIX. 

THEORY  OF  LIMITS 170 

SYMMETRY , 175 

THEOREMS  ON  SYMMETRY...  .  177 


UNIVERSITY 


INTRODUCTION. 


GEOMETRY  is  highly  important,  and  growing  in  importance 
as  a  branch  of  mathematical  study.  This  is  not  only  true  for 
mathematicians ;  but,  what  is  extremely  interesting  to  teachers, 
to  practical  men  also.  The  vast  development  of  machinery,  of 
steam-power,  of  the  applications  of  electricity,  bring  chapters 
of  mathematical  science  into  every-day  use  which  have  long 
been  employed  in  physical  investigation,  but  never  before  in 
the  arts  of  life;  and  in  the  construction  of  all  kinds  of 
machinery  and  industrial  devices  the  geometrical  representa- 
I  tion  to  the  eye  is  of  far  more  immediate  and  practical  value 
than  abstract  calculations.  The  men  who  can  draw  are  rapidly 
gaining  on  those  who  can  merely  calculate.  In  such  matters 
as  statistics  the  old  processes  are  even  reversed.  We  have  not 
only  an  application  of  arithmetic  to  geometry,  but  geometrical 
representations  of  arithmetical  results ;  there  is  not  merely  an 
algebraic  geometry,  but  a  graphic  algebra. 

In  manual  training  schools  geometry  is  fully  as  important  a 
branch  of  mathematics  as  arithmetic,  even  for  the  future 
mechanic.  The  great  defect  in  American  mathematical  train- 
ing has  been  that  arithmetic  and  algebra  have  been  too  much 
favored  as  against  geometry.  Teachers  have  delayed,  and 
still  delay,  presenting  even  the  elements  of  geometry  till  a 
great  deal  of  algebra  has  been  mastered;  the  meagre  facts 
which  must  be  stated  before  even  mensuration  can  be  intelli- 
gently treated  have  been  reduced  to  the  smallest  compass  and 
the  most  mechanical  shape ;  and  it  is  only  because  we  confuse 
the  difficulty  of  the  subject  with  our  stupid  ways  of  teaching 
it  that  we  tolerate  the  geometrical  ignorance  of  our  pupils. 

ix 


X  INTRODUCTION. 

There  are  skilful  mathematicians  who  are  unaware  how 
much  better  geometers  early  training  would  have  made  them, 
if  the  geometrical  side  of  things  had  had  fair  play  in  their 
education.  But  a  reform  is  impending;  and  Mr.  Hopkins's 
text-book  here  presented  is  intended  to  promote  it.  I  desire 
to  call  the  attention  of  all  earnest  and  progressive  teachers  to 
the  heuristic  method  as  here  expounded. 

The  word  heuristic  is  derived  from  the  Greek ;  it  means  the 
method  of  discovery.  Inventional  is  another  word  which  has 
been  somewhat  similarly  used;  but,  I  think,  in  a  narrower 
meaning. 

In  mathematics  the  "  heuristic"  is  the  same  as  the  "  develop- 
ment "  method ;  that  is,  the  method  by  which  the  pupil  is  led 
to  see  the  theorems  and  their  demonstrations  for  himself. 

In  attempting  to  use  this  book,  the  ordinary  laws  of  good 
teaching  must  be  followed.  Consequently  the  pupils,  however 
mature,  must  possess  all  the  prior  qualifications ;  they  must  be 
intelligent  as  to  the  subject-matter.  The  greatest  difficulty  in 
now  teaching  geometry  by  any  method  lies  in  that  neglect  of  the 
elements  to  which  I  have  before  alluded.  If  a  pupil  reaches 
the  age  of  fifteen  (as  the  pupils  of  the  Massachusetts  grammar 
schools  are  said  to  do)  without  thorough  and  systematic  train- 
ing in  the  elements  of  the  subject,  nothing  remains  but  to  pre- 
fix a  course  of  instruction  in  these  elements  to  the  study  of 
demonstrative  geometry.  A  very  skilful  and  celebrated  teacher 
was  about  1875  mentioned  to  me  by  his  pupils  as  using  the 
heuristic  method;  and  I  at  once  wrote  him  to  inquire  how  he 
presented  the  elements.  His  answer  was  that  about  six  weeks 
were  spent  in  working  up,  orally,  the  doctrine  of  form;  that 
important  part  of  geometry  whose  results  are  contained  in  the 
definitions  and  other  preliminary  matter.  This  was  in  America, 
with  pupils  who  had  passed  the  grammar  school  without  learn- 
ing these  elements ;  and  represents  what  seems  to  the  writer 
to  be  the  minimum  of  attention  to  be  given  to  this  part  of  the 
subject. 


INTRODUCTION.  xi 

On  the  other  hand,  the  Austrian  higher  schools  spend  four 
years  (one  and  one-half  to  two  hours  weekly)  in  that  empirical 
form  of  geometiy  in  which  attention  is  given  rather  to  acquir- 
ing a  knowledge  and  practical  use  of  the  subject  (both  plane 
and  solid)  than  to  its  logic  as  a  strictly  scientific  branch  of 
study.  These  boys  are  from  eleven  to  fifteen  years  old,  on  the 
average,  and  correspond  to  our  grammar  school  pupils  in  age 
and  general  maturity.  Their  instruction,  so  far  as  I  can  judge 
from  the  text-books  and  the  books  of  direction  to  the  teachers, 
is  almost  entirely  heuristic  in  character ;  and  a  former  pupil 
of  mine  of  Austrian  birth  (who  is  now  an  eminent  American 
professor  of  an  ancient  language)  confirmed  this,  and  assured 
me  that  then  (twenty  years  ago)  this  method  was  actually 
employed  in  his  training. 

In  the  earlier  stages  of  such  a  course,  immediate  inspection 
shows  the  simpler  geometrical  truths  with  full  conviction  and 
ready  acceptance  j  precisely  as  the  pupil  learns  that  seven  times 
nine  equals  nine  times  seven,  not  by  algebraic  demonstration 
(which  involves  several  steps),  but  by  practical  experience. 
Mathematicians  in  the  higher  branches  make  great  use  of 
intuitional  proofs,  supplemented  if  need  be  by  logical  demon- 
stration; and  in  the  lower  the  same  law  holds  good  on  the 
geometrical  side  as  well  as  on  the  arithmetical. 

The  beginner  in  geometry  should  not  at  first  be  required  to 
perform  all  the  geometrical  operations  at  once,  nor  with  great 
rapidity.  The  careful  series  of  objective  illustrations  of  geo- 
metrical form,  combined  with  practical  exercises  and  simple 
reasonings,  cannot,  under  our  present  programme,  receive  all 
the  expansion  which  it  has  in  Austria.  Our  teachers  are  com- 
pelled to  deal  with  their  pupils  in  a  short  time,  and  to  teach 
them  to  geometrize  in  comparatively  few  lessons ;  and  so  the 
time  to  practically  carry  out  the  heuristic  method  must  be 
obtained  by  economizing  opportunities. 

But  the  old-fashioned  method  of  memorizing  the  whole  book, 
definitions,  propositions,  corollaries,  scholia,  even  the  numbers 


Xll  IN  TE  OD  UCTION. 

which  were  prefixed  to  these  truths  to  indicate  their  order,  is 
happily  on  the  decline.  It  is  a  wasteful  method,  and  commu- 
nicates far  more  information  than  can  be  permanent  or  useful; 
and  it  lays  a  heavy  burden  on  the  memory.  When  the  heu- 
ristic method  takes  its  place,  the  pupil  will  first  of  all  be 
brought,  by  successive  steps  of  abstraction,  from  the  study  of 
geometrical  models,  to  the  idea  of  geometrical  solids,  surfaces, 
lines,  points ;  they  will  then  be  taught  the  generation  of  lines 
by  the  motion  of  points,  the  various  classes  of  lines,  the  sim- 
plest figures  bounded  by  straight  lines,  triangle,  square,  rec- 
tangle, rhomboid,  trapezoid,  and  trapezium.  The  analysis 
should  be  directed  to  the  cube,  the  other  four  regular  solids, 
the  ordinary  solids  bounded  by  straight  lines.  From  these 
the  idea  of  plane  angles  in  their  various  classes  should  be 
obtained;  the  difference  and  relationship  between  plane  and 
diedral  angles  should  be  insisted  upon.  In  what  does  an  angle 
011  the  blackboard  differ  from  that  formed  when  a  book  or  a 
door  is  opened  ? 

The  doctrine  of  form  should  be  extended  somewhat  beyond 
the  things  mentioned  in  plane  geometry;  the  reciprocal  relation 
between  the  cube  and  the  octaedron  (each  has  as  many  faces 
as  the  other  has  vertices ;  the  number  of  their  edges  is  equal) 
should  be  pointed  out  on  the  model  and  exemplified  by  draw- 
ing one  inscribed  in  the  other.  Similar  relations  exist  between 
the  dodecaedron  and  icosaedron ;  and  two  tetraedrons.  Again, 
in  analyzing  all  convex  solids  bounded  by  plane  faces  the  fact 
should  be  brought  out  that  E  +  2  =  F+  V',  that  is,  the  cube 
has  12  edges,  6  faces,  8  vertices;  12  +  2  =  6  +  8. 

A  skilful  object  teacher,  with  a  few  models,  can  thus  readily 
develop  a  great  many  interesting  truths,  and  thus  prepare  the 
young  people's  minds  for  geometry;  a  few  weeks'  time  can 
be  well  spent  on  the  definitions. 

Euclid's  text-book,  which  is  on  a  different  plan  from  the 
modern  books,  begins  rather  more  gradually;  and  our  best 
teachers  imitate  him  in  going  more  slowly  over  the  early  prop- 


IN  TR  OD  UCTION.  xiii 

ositions.  It  seems  to  me  that  the  models  bounded  by  curved 
surfaces  (cylinder,  cone,  sphere,  frustum  of  a  cone,  and  the 
parts  of  a  cone  bounded  by  the  sections,  as  well  as  the  ellip- 
soid) ought  to  be  exhibited  and  so  far  analyzed  as  to  furnish 
material  for  the  definitions  of  straight  lines  and  plane  surfaces, 
in  distinction  to  curved  lines  and  surfaces.  And  I  would  go  a 
step  farther;  and  show  the  difference  between  the  ruled  sur- 
faces (like  the  cylinder  and  the  cone)  and  those  in  which  no 
straight  lines  can  be  drawn. 

In  a  word,  I  would  lay  the  foundation  for  solid  geometry 
(which  is  the  practical  form  of  the  science)  along  with  plane, 
which  is  a  mere  abstraction  from  the  other.  The  introduction 
to  demonstration  may  well  be  combined  with  additional  study 
of  form.  Let  the  pupils,  for  instance,  be  taught  to  make  tri- 
angles; then  to  measure  the  sides  and  angles  of  those  they 
make;  then  to  classify  their  triangles  into  equilateral,  isos- 
celes, scalene;  equiangular,  triangles  with  two  equal  angles, 
with  no  equal  angles ;  then  to  set  down  a  few  theorems  they 
may  expect  to  prove.  This  is  no  doubt  the  way  the  Egyptian 
priests  built  up  the  propositions  into  which,  as  the  foundations 
of  a  secret  society,  they  initiated  Pythagoras.  Eight  angles, 
vertical  angles,  exterior  angles,  can  now  be  brought  in  natu- 
rally; then  triangles  with  a  short  side  and  two  long  ones; 
then  the  idea  of  parallels.  Any  bright  teacher  who  is  pos- 
sessed with  the  desire  of  training  pupils  to  think  geometri- 
cally will  naturally  fall  into  the  line  of  thinking  needed;  but 
it  all  takes  time. 

I  estimate  that,  when  the  definitions  and  those  matters 
which  make  up  the  first  book  of  our  ordinary  geometries  have 
been  fully  mastered,  half  the  work  of  the  course  in  plane 
geometry  has  been  accomplished;  and  the  teacher  must  bear  in 
mind  that  if  this  is  well  done,  the  pupil  has  been  trained,  not 
crammed,  to  a  very  high  point  in  the  subject. 

The  modern  writers  on  pedagogy  lay  down  the  principle 
that  interest  is  the  end  of  teaching,  and  not  merely  the  means. 


XIV  INTRODUCTION. 

If  you  have  brought  your  pupil  to  enter  upon  mathematics 
with  an  earnest  desire  to  accomplish  the  work  in  it,  you  have 
done  more  than  if  you  merely  cram  it  into  him.  In  the  col- 
leges it  is  well  known  that  the  great  mass  of  students  do  com- 
paratively little  in  the  science.  When  they  have  the  choice, 
they  prefer  as  a  rule  to  take  something  easier.  Just  so  most 
boys  never  accomplish  much  in  their  athletics.  The  fashion 
has  come  about  that  about  a  quarter  of  our  students  play 
ball,  more  or  less,  and  the  rest  look  on ;  simply  because  their 
muscles  have  not  been  trained  to  the  work.  Now  what  we  want 
in  mathematics  is  precisely  analogous, — training  of  the  mathe- 
matical muscles.  The  German  use  of  "  gymnasium  "  as  the  name 
for  those  high  schools  which  do  the  work  preparatory  to  the  uni- 
versities —  schools  which  accomplish  nearly  as  much  at  the  age 
of  nineteen  or  twenty  as  our  colleges  do  three  years  later  on  — 
is  in  the  same  line  of  thinking.  Let  us  make  our  high  schools 
true  gymnasia;  and  a  good  subject  to  begin  with  is  geometry. 

What,  then,  is  the  true  exercise  of  the  mathematical  powers  ? 
(I  had  almost  said  of  the  mathematical  muscles.)  Grube's 
numerical  analysis  of  the  numbers  from  1  to  100  is  the  heuristic 
method  in  arithmetic;  the  pupils  are  led  on  by  the  teacher  to 
analyze  and  find  out  arithmetical  truths  for  themselves.  It  is 
much  to  be  wished,  as  Pestalozzi  taught,  that  form  might  be 
thoroughly  taught  in  the  primary  school;  Steiner,  whose 
geometrical  ideas  lie  nearly  at  the  foundation  of  modern 
developments,  was  a  pupil  and  teacher  of  Pestalozzi's  own 
school  in  Switzerland.  But  if  this  is  not  yet  practicable,  let 
the  pupil  take  up  geometry  from  the  beginning  at  a  later  age; 
and  really  do  fundamental  and  thorough  work. 

The  teacher,  in  using  the  heuristic  method,  must  be  very 
careful  of  his  foundations;  he  must  develop,  and  thoroughly 
develop,  his  definitions,  instead  of  giving  them  out  as  a  task. 
The  first  demonstrations  must  be  slowly  and  carefully  done ; 
neither  the  Egyptian  priests  nor  Pythagoras  discovered  the 
early  theorems  all  at  once,  or  in  any  short  time. 


INTRODUCTION.  XV 

The  hill  of  mathematical  knowledge  is  high  and  steep,  and 
few  go  up  on  it  any  great  distance.  There  are  three  ways 
of  ascending.  The  pupil  may  be  carried  up  in  a  wagon;  this 
is  tolerably  easy,  and  there  is  some  gain  of  fresh  air  and  a  wide 
prospect.  It  is  like  the  committal  to  memory  of  a  text- 
book. He  may  be  dragged  up  as  tourists  sometimes  ascend 
the  Alps;  tied  to  a  rope  with  a  strong  guide  at  either  end. 
This  is  still  more  beneficial ;  but  the  best  way  of  all  is  to  be 
trained  to  walk  up  for  himself.  At  first,  the  trainer  must 
content  himself  with  gradual  development  of  the  muscles,  and 
little  apparent  ascent;  but  the  continual  exercise  finally  enables 
him  to  reach  a  great  height  with  comparative  ease. 

In  preparing  pupils  to  pass  college  examinations  the  heuristic 
method  has  a  great  practical  value.  Once  the  method  of  mak- 
ing a  demonstration  is  known  and  the  few  data  upon  which 
the  questionsxiepend  are  learned,  the  student  who  can  demon- 
strate originally  has  a  great  advantage  over  him  who  relies 
more  largely  on  his  memory.  If  the  latter  forgets,  he  is  lost ; 
but  the  former  can  originate  when  memory  fails,  and  if  he  has 
been  well  trained  is  far  less  liable  to  lose  his  presence  of  mind. 

I  congratulate  Professor  Hopkins  on  the  spirit  with  which 
he  has  undertaken  to  fill  a  gap  in  our  geometrical  literature ; 
and  I  hope  the  book  will  have  the  success  which  it  deserves. 

TRUMAN  HENRY  SAFFORD. 

DEPARTMENT  OF  ASTRONOMY, 

WILLIAMS  COLLEGE, 

August,  1891. 


"HE 

iVERSJTY '  \ 

, 

y 
BOOK  I. 


PLANE    GEOMETRY. 


1.  Space  is  indefinite  extension  in  every  direction. 

2.  A  material  substance  is  anything,  large  or  small,  solid, 
liquid,  or  aeriform,  visible  or  invisible,  that  occupies  a  portion 
of  space. 

3.  It  therefore  follows  that  material  substances  have  limited 
extension  in  every  direction. 

4.  For  purposes  of  measurement,  extension  in  three  direc- 
tions only  are  considered,  called,  respectively,  length,  breadth 
(or  width),  and  thickness;   they  are  also  called  collectively 
dimensions. 

5.  Magnitude,  in  general,  means  size,  and  is  applied  to  any- 
thing of  which  greater  or  less  can  be  predicated,  as   time, 
weight,  distance,  etc.;  a  geometrical  magnitude  is  that  which 
has  one  or  more  of  the  three  dimensions. 

6.  A  geometrical  point  has  position  merely;   i.e.  it  has  no 
magnitude. 

The  dots  made  by  pencil  and  crayon  are  called  points,  but 
they  are  really  small  substances  used  to  indicate  to  the  eye 
the  location  of  the  geometrical  point. 

7.  A  geometrical  line  has  only  one  dimension ;  i.e.  length. 
The  lines  made  by  pencil  and  crayon  are  substances,  and 

1 


2  PLANE  GEOMETRY. 

may  be  called  physical  lines  which  serve  to  show  the  position 
of  the  geometrical  lines. 

8.  A  straight  line  is  one  that  lies  evenly  between  its  ex- 
treme points. 

This  is  the  definition  as  given  by  Euclid.  The  majority 
of  modern  geometers,  however,  have  substituted  the  following 
as  stated  by  Newcomb ;  viz.  : 

"  A  straight  line  is  one  which  has  the  same  direction  through- 
out its  whole  length." 

Each  is  designed  to  express  the  idea  of  straightness,  and  not 
to  convey  it,  for  it  is  assumed  that  the  idea  already  exists  in 
the  pupil's  mind  prior  to  the  beginning  of  this  study. 

9.  A  curved  line,  or  simply  curve,  is  one  no  part  of  which 
is  straight. 

10.  Material   substances    have    one  .  or   more    faces   which 
separate  them  from  the  rest  of  space.     These  faces  are  called 
surfaces,  and  have,  obviously,  only  two  dimensions ;  i.e.  length 
and  breadth. 

11.  The  surface  considered  apart  from  the  substance  is  called 
a  geometrical  surface. 

12.  A  plane  is  a  geometrical  surface  such  that  if  any  two 
points  in  it  be  selected  at  random,  the  straight  line  joining 
them  will  lie  wholly  in  that  surface. 

13.  A  curved  surface  is  a  geometrical  surface  no  portion  of 
which  is  a  plane. 

14.  A  physical  solid  is  the  material  composing  it,  and  which 
we  perceive  through  the  medium  of  the   senses ;   while  the 
geometrical  solid  is  the  space,  simply,  which  the  physical  solid 
occupies. 

15.  A  geometrical  figure  is  the  term  applied  to  combinations 
of  points,  lines,  and  surfaces,  when  reference  is  had  to  their 
form  or  outline  simply. 


PLANE  GEOMETRY.  8 

16.  A  plane  figure  is  one  whose  points  and  lines  all  lie  in 
the  same  plane. 

17.  "A  plane  rectilineal  angle   is   the   inclination  of   two 
straight  lines  to  one  another,  which  meet  together,  but  are 
not  in  the  same  straight  line."  —  EUCLID. 

"  An  angle  is  a  figure  formed  by  two  straight  lines  drawn 
from  the  same  point."  —  CHAUVENET. 

"When  two  straight  liaes  meet  together,  their  mutual  in- 
clination, or  degree  of  opening,  is  called  an  angle"  —  LOOMIS. 

18.  The  lines  which  form  an  angle  are  called  the  sides  of 
the  angle,  and  the  point  from  which  they  are  drawn  is  called 
the  vertex  of  the  angle. 

19.  When  two  plane  angles  have  the  same  vertex  and  a 
common  side,  neither  angle  being  a  part  of  the  other,  they  are 
said  to  be  adjacent  angles. 

20.  When  two  angles  have  the  same  vertex  and  the  sides 
of  one  are  the  extensions  of  the  sides  of  the  other,  they  are 
called  vertical  angles. 

21.  An  angle  is  named  by  a  letter  or  number  placed  at  its 
vertex.     If,  however,  there  are  two  or  more  angles  with  the 
same  vertex,  other  letters  are  placed  at  the  extremities  of 
their  sides,  and  the  three  letters  are  used  to  name  the  angle, 
the  letter  at  the  vertex  always  coming  between  the  other  two. 

22.  Let  us  consider  the  point  B,  in  the  straight  line  AC, 
a  pivot,  and  BD  another  starting  from  the  position  BC,  and 

•j-v    D  T» 


A  B  C     A  B  C 

Fig.  I.  Fig.  H. 

turning  about  B,  keeping  always  in      

the  same  plane.     It  is  evident  that,     A 

as  soon  as  it  has  started,  it  forms 

two  angles  with  the  line  AC,  of  which  DBC,  Fig.  I.,  is  the 


4  PLANE  GEOMETRY. 

smaller.  If  it  continue  to  revolve,  however,  it  will  finally 
reach  a  position  as  DB,  Fig.  II.,  in  which  the  angle  DEC  is 
the  larger.  Hence,  in  passing  from  the  first  position  to  the 
second,  it  must  have  reached  a  position,  Fig.  III.,  where  the 
two  angles  DEC  and  DBA  were  equal. 

23.  Hence,  when  one  straight  line  meets  another  so  as  to 
form  equal  adjacent  angles,  each  of  the  angles  is  called  a  right 
angle,  and  the  lines  are  said  to  be  perpendicular  to  each  other. 

24.  It  is  also  evident  that  the  sum  of  the  angles  formed 
by  any  one  position  of  the  line  ED  is  equal  to  the  sum  of  the 
angles  formed  by  any  other  position ;  for  what  is  taken  from 
one  angle  by  the  revolution  of  the  line  ED  is  added  to  the 
other. 

25.  Hence,  when  one  straight  line  meets  another  so  as  to 
form  two  angles,  the  sum  of  these  two  angles  equals  two 
right  angles. 

26.  An   angle  that  is  less  than  a  right  angle  is  called  an 
acute  angle. 

27.  An  angle  that  is  greater  than  one  right  angle  and  less 
than  two  is  called  an  obtuse  angle. 

28.  Both  acute  and  obtuse  angles  are  designated  as  oblique 
angles  as  contrasted  with  right,  angles, 

29.  A  straight  angle  is  a  term  recently  adopted  by  promi- 
nent English  and  German  mathematicians  to  express,  by  a 
single  unit,  the  sum  of  two  right,  angles. 

30.  When  the  sum  of  two  angles  is  equal  to  a  straight 
angle,  they  are   said  to  be   supplementary  >   i.e.  each  is  the 
supplement  of  the  other. 

31.  When  the  sum   of  two   angles  is  equal  to  one  right 
angle,  they  are  said  to  be  compl&nientq/ry ;  i.e.  each  is  the 
complement  of  the  other. 


UN' 

PLANE  GEOMETRY.  5 

32.  It  is  evident  from  22  and  23  that  when  one  straight 
line  meets  another  so  as  to  form  two  angles,  these  angles  are 
supplementary. 

33.  Two  straight  lines  are  said  to  be  parallel  when,  lying  in 
the  same  plane,  and  extended  indefinitely  both  ways,  they 
do  not  meet  each  other. 

34.  As  we  have  before  conceived  a  line  (22)  to  move,  so 
we  may  conceive  one  geometrical  magnitude  to  be  applied  to 
another   for   the   purpose   of   comparison.     If  they   coincide, 
point  for  point,  they  are  said  to  be  equal. 

35.  Thus,  if  two  angles  can  be  so  placed  that  their  vertices 
coincide  in  position  and  their  sides  in  direction,  two  and  two, 
the  angles  must  be  equal. 

36.  Conversely,  if  two  equal  angles  be  conceived  to  be  so 
placed,  one  upon  the  other,  that  their  vertices  and  one  pair 
of   sides  coincide  respectively,  then  the  other  pair  of  sides 
must  also  coincide,  otherwise  one  angle  would  be  greater  than 
the  other. 

37.  Geometrical  magnitudes  are  geometrical  lines,  angles,  sur- 
faces, and  solids. 

38.  We  shall  have  occasion  to  express   the   addition   and 
subtraction  of  geometrical  magnitudes,  as  well  as  the  multi- 
plication and  division  of  these  magnitudes  by  numbers. 

39.  For  example,  the  sum  of  the  two  lines  AB  and  CD  is 
obtained  by  conceiving  them  to  be  placed  so  as  to  form  one 
continuous  straight  line  as  HK.  .  R 
Similarly,  the  difference  of  two 

lines  is  obtained  by  cutting  off 

from  the  larger  a  line  equal  to    — —  — -r- —  — z 

the  smaller.      Similarly,  HKN 

represents  the  sum  of  the  two  angles  A  and  B.     To  multiply 

a  line  by  a  number  is  to  add  it  to  itself  the  required  number 

of  times.     (See  above.) 


6  PLANE  GEOMETRY. 

To  divide  a  line  by  a  number  is  to  conceive  the  line  to  be 
divided  into  the  required  number  of  equal  parts. 


HI 

The  same  is  true  of  other  geometrical  magnitudes.     (Illus- 
trations of  each  should  be  given.) 

40.   An  axiom  is  a  truth  that  needs  no  argument ;  i.e.  the 
mere  statement  of  it  makes  it  apparent,  e.g. : 

GENERAL  AXIOMS. 

(I.  The  whole  of  anything  is  greater  than  any  one  of  its  parts. 

•  II.  The  whole  of  anything  is  equal  to  the  sum  of  all  its  parts. 
/Til.  Quantities  which  are  respectively  equal  to  the  same  or 
equal  quantities  are  equal  to  each  other. 

7"lV.   Quantities  which  are  respectively  halves  of  the  same  or 
equal  quantities  are  equal  to  each  other. 

V.  Quantities  which  are  respectively  doubles  of  the  same 
or  equal  quantities  are  equal  to  each  other. 

VI.  If  equal  quantities  be  added  to  equal  quantities,  the 
sums  are  equal. 

VII.  If  equal  quantities  be  subtracted  from  equal  quantities, 
the  remaining  quantities  are  equal, 

VIII.  If   equal   quantities   be   multiplied  by  the  same  or 
equal  quantities,  the  products  are  equal. 


PLANE  GEOMETRY.  7 

/      IX.   If  equal  quantities  be  divided  by  the  same  or  equal 
quantities,  the  quotients  are  equal. 

X.  If  equal  quantities  be  either  added  to  or  subtracted  from 
Lunequal  quantities,  the  results  will  be  unequal. 

XI.  If  equal  quantities  be  either  multiplied  or  divided  by 
unequals,  the  results  will  be  unequal. 

41.  The   results   obtained  by  the   addition  to,  subtraction 
from,  multiplication  or  division  of,  unequals  by  unequals  are 
indeterminate  with  one  exception.    The  pupil  should  ascertain 
for  himself  this  exception. 

42.  Particular  axioms. 

XII.  Between   two   points   only  one   straight  line  can  be 
^     drawn  5  or  if  others  are  drawn,  they  must  coincide. 

TXIII.    A  straight  line  is  the  shortest  of  all  possible  lines 
Connecting  two  points, 

XIY.   Conversely,  the  shortest  line  between  two  points  is 
a  straight  line. 

'""XV.   If  two  straight  lines  have  two  points  in  common,  they 
will  coincide  however  far  extended. 

XVI.  Two  straight  lines  can  intersect  in  only  one  point. 

XVII.  In  one  direction  from  a  point  only  one  straight  line 
can  be  drawn ;  or  if  more  be  drawn,  they  must  coincide, 

XVIII.  Through   a  given     , 

point  (as  P)    only   one  line     A  P  B 

(as  AB)  can  be  drawn  par-     •- —  — — 

allel  to  another  line  (as  CD) ; 
r  if  others  are  drawn,  they  must  coincide, 

XIX.  If  a  line  makes 
an  angle  with  one  of  two 
parallel  lines,  it  will  in- 
tersect the  other  if  sufficiently  extended. 

XX.  The  extension  or  shortening  of  the  sides  of  an  angle 
does  not  change  the  magnitude  of  the  angle. 


8  PLANE  GEOMETRY. 

43.  A   theorem,  is   a  truth  which  is   made   apparent   by  a 
course  of  reasoning  or  argument.     This  argument  is  called  a 
demonstration. 

Every  theorem  consists  of  two  distinct  parts,  either  ex- 
pressed or  implied;  viz.  the  hypothesis  and  conclusion.  The 
conclusion  is  the  part  to  be  proven,  and  the  demonstration  is 
undertaken  only  upon  the  ready  granting  of  the  conditions 
expressed  in  hypothesis  ;  e.g. : 

Hyp.   If  two  parallel  lines  be  crossed  by  a  transversal, 

Con.   the  alternate  interior  angles  are  equal. 

44.  In  demonstrating  the  theorems  in  this  book  the  pupil 
should  first  analyze  the  theorem  and  write  it  after  the  above 
model.     For  instance,  let  us  analyze  the  following  theorem; 
viz. : 

A  perpendicular  measures  the  shortest  distance  from  a  point 
to  a  straight  line. 

Now  this  theorem,  analyzed  and  written  according  to  our 
model,  would  read  as  follows : 

Hyp.  If  from  a  given  point  to  a  given  straight  line  a  per- 
pendicular and  other  lines  be  drawn, 

Con.   the  perpendicular  will  be  the  shortest  one  of  those  lines. 

45.  The  converse  of  a  theorem  is  another  theorem  in  which 
the  hypothesis  becomes   the  conclusion,  and  conclusion  the 
hypothesis.     For  example,  the  converse  of  the  theorem  men- 
tioned in  Sect.  43  would  read  as  follows ;  viz. : 

Hyp.  If  two  straight  lines  in  the  same  plane  be  crossed  by 
a  transversal  so  as  to  make  the  alternate  interior  angles  equal, 

Con.   these  two  straight  lines  will  be  parallel. 

The  converse  of  most  of  the  theorems  in  this  work"will  be 
left  for  the  pupil  to  state. 

46.  A  problem,  in  geometry,  is  the  required  construction  of 
a  geometrical  figure  from  stated  conditions  or  data ;  e.g. : 

It  is  required  to  construct  the  triangle  which  has  for  two 


PLANE  GEOMETRY.  9 

of  its  sides  AB  and  CD,  and  the  angle  H  included  between 
these  two  sides. 

ZL  J3 

A  postulate  is  a  self-evident  problem, 
or  a  construction  to  the  possibility 
of  which  assent  may  be  demanded  or 
challenged  without  argument  or  evi- 
dence. 

(Both  theorems  and  problems  are 
commonly  designated  as  propositions.) 

47.  Before  we    can    accomplish  the  demonstration  of    a 
theorem    in    geometry,   the    following    postulates    must    be 
granted ;  viz. : 

I.  A  straight  line  can  be  drawn  from  one  point  to  any 
other  point. 

II.  A  straight  line  can  be  extended  to  any  length,  or  ter- 
minated at  any  point. 

III.  A  circle  may  be  described  about  any  point  as  a  centre 
and  with  any  radius. 

IV.  Geometrical   magnitudes   of   the  same  kind  may  be 
added,  subtracted,  multiplied,  and  divided. 

V.  A  geometrical   figure   may  be   conceived  as  moved  at 
pleasure  without  changing  its  size  or  shape. 

48.  A  postulate  is  also  used  to  designate  the  first  step  in  the 
demonstration  whereby  certain  conditions  or  data  are  demanded 
as  fulfilled,  or  admitted  as  true,  for  the  basis  of  the  argument, 
and  which  begins   with  "Let,"  etc.,  or  "Let  it  be  granted 
that,"  etc.     It  really  demands  assent    ~ 

to  the  general  conditions  implied  in     ^%SX. 

the  hypothesis,  with  special  reference 

to  a  particular  representative  diagram 

or  figure ;  e.g.  in  beginning  the  demon- 

stration  of  the  theorem  given  in  Sect.  ^x 

43,  we  should  say :  "Let  AB  and  HK 

be  two  parallel  straight  lines  crossed  by  the  transversal  C7Z>." 


10 


PLANE  GEOMETRY. 


This  is  the  postulate ;  i.e.  it  matters  not  whether  AB  and 
HK  are  actually  straight  or  actually  parallel ;  they  stand  for 
straight  and  parallel  lines,,  and  the  argument  is  just  as  con- 
clusive when  based  upon  their  supposed  parallelism,  as  it 
would  be  if  we  had  positive  knowledge  that  those  two  iden- 
tical lines  were  parallel.. 

49.  The  demonstration  of  the  following  theorems,  should  be 
written  by  the  pupils,  with  an  occasional  oral  exercise  partici- 
pated in  by  the  entire  class,  each  member  in  turn  contributing 
a  single  link  in  the  chain  of  argument. 

50.  In  order  to  save  time  for  the  pupil  in  writing  and  the 
instructor  in  correcting,  the  author,  having  used  them,  recom- 
mends the  use  of  the  following  list  of  symbols  and  abbrevia- 
tions, as  well  as  such  others  as  the  instructor  and  pupils  may 
agree  upon : 


51.  SYMBOLS, 

4-  .  .  .  plus. 
—  ...  minus. 
X   ...  multiplied  by. 
=  .  .  .  equals,  or  is  equal  to. 
.'.  .  .  .  therefore,  or  hence. 
>  ...  is  greater  than. 
<  ...  is  less  than. 
=  or  X,  equivalent  to. 
O  .  .  .  circle. 
©   ...  circles. 

11    ...  parallel. 

IPs    .  .  parallels. 


/. angle. 

A  .  .  .  .  .  angles, 

rt.  Z  or  [R_  .  .  right  angle, 

rt.  A  or  [R_'S  .  right  angles. 

A triangle. 

A triangles. 

rt.  A right  triangle. 

rt.  A  .  .  .  .  »  right  triangles. 

_L perpendicular. 

Js perpendiculars. 

O parallelogram. 

£17 parallelograms. 


PLANE  GEOMETRY. 


11 


52. 


ABBKEVIATIONS, 


Adj adjacent. 

Alt alternate. 

Ax axiom. 

Comp complementary, 

Con conclusion. 

Cons construction. 

Def definition. 

Dem demonstration. 

Dist distance. 

Ext exterior. 

Hyp.     ....  hypothesis. 

Iden identical. 

Q.E.D.  .  .  .  Quod 
Q.E.F.  .  .  .  Quod 


Int interior. 

Line straight  line. 

Opp opposite. 

Post postulate. 

Prob problem. 

Pt point. 

Quad quadrilateral. 

St straight. 

Sug suggestion. 

Sup supplementary. 

Trans transversal. 

Vert vertical. 

erat  demonstrandum, 
erat  faciendum. 


53.  The  last  two  expressions  are  in  Latin,  and  mean  respec- 
tively, "which  was  to  be  demonstrated"  or  "proven,"  and 
"which  was  to  be  performed"  or  "done."      The  former  is 
placed  at  the  close  of  the  demonstration  of  every  theorem  to 
indicate  that  the  required  proof  has  been  completed ;.  while  the 
latter  is  placed  at  the  close  of  the  work  of  every  problem  to 
indicate  that  the  required  construction  has  been  completed. 

54.  The  pupil  must  remember  that  every  statement  in  geo- 
metrical demonstration  must  be  "  backed  up  "  or  substantiated 
by   giving   as   authority,  definitions,  axioms,   and  previously 
established  truths,  and  every  unsupported  statement,  whether 
from  instructor  or  fellow-pupil,  should  be  promptly  challenged. 

55.  Complete   demonstrations  are  given  of  only  a  few  of 
the  following  theorems,  and  those  solely  for  the.  purpose  of 
giving  the  pupil  a  start  when  encountering  a  new-  subject, 
or  in  places  where  the  pupil's  time  would  not  be  spent  to 
advantage  in  his  helpless  work ;  i.e.  where  the  average  pupil 
would  be  at  a  loss  to  know  how  to  go  at  the  subject.     In 


12  PLANE  GEOMETRY. 

other  places  occasional  suggestions  are  made  for  the  purpose 
of  giving  the  pupil  a  start  without  loss  of  time.  In  only  a 
few  cases  have  any  diagrams  been  given,  the  author  thinking 
it  best  to  leave  them  as  a  part  of  the  pupil's  exercise.  And 
right  here  he  wishes  to  emphasize  this  caution  to  both  in- 
structor and  pupil,  based  on  several  years'  experience  in  the 
class-room ;  viz. :  Always  employ  the  most  unfavorable  diagram. 
Care  in  this  particular  will  save  many  an  inadvertent  error, 
and  prevent  the  assumption  of  conditions  unwarranted  by  the 
hypothesis. 

THEOREMS. 

56.   Hyp.   If  one  straight  line  intersects  another, 
Con.   the  vertical  angles  are  equal. 


B 


Post.   Let  AH  and  BD  be  two  straight  lines  intersecting 
each  other  at  point  C. 

We  are  to  prove  Z  ACD  =  Z  BCH  (I.) 

and                           ZACB  =  Z.DCH.  (II.) 

Dem.    Z  ACD  +  Z  ACS  =  2  rt.  A.  (Sect.  32.) 

ZBCH  +  ZACB  =  2rt.  A  (Same  reason.) 

.-.  /.ACD  +  /.ACB  =  /.BCH+/.ACB.   (Axiom  III.) 

ZACB=  ^ACB.             (Iden.) 

/.  Z  ACD                 =  Z  BCH.  (Axiom  VII.) 

Q.E.D. 

The  pupil  should  now  employ  a  similar  method  and  demon- 
strate (II.)  and  the  following  easy  ones : 


PLANE  GEOMETRY.  13 

57.  If  two  straight  lines  intersect  each  other,  the  sum  of 
the  four  angles  thus  formed  equals  four  right  angles. 

58.  The  sum  of  all  the  angles  formed  at  one  point  on  the 
same  side  of  a  straight  line  equals  two  right  angles. 

59.  The  sum  of  all  the  angles  formed  by  any  number  of 
straight  lines  meeting  at  one  point  equals  four  right  angles. 

Sug.   Extend  one  of  the  lines. 

Remark.  In  the  demonstration  of  most  theorems  it  be- 
comes necessary  to  draw  one  or  more  auxiliary  lines  in  addi- 
tion to  those  involved  in  the  original  figure.  These  are  called, 
technically,  construction  lines,  and  should  always  be  dotted,  in 
order  that  they  may  readily  be  distinguished  from  the  given 
lines,  or  those  comprising  the  original  figure. 

60.  If  one  of  the  four  angles  formed  by  the  intersection 
of  two  straight  lines  is  a  right  angle,  the  other  three  angles 
are  also  right  angles. 

61.  All  right  angles  are  equal. 

62.  If  two  angles  be  equal,  the  complements  of  those  angles 
are  also  equal. 

63.  If  two  angles  be  equal,  the  supplements  of  those  angles 
are  also  equal. 

64.  If  two  supplementary  adjacent  angles  be  bisected,  the 
bisecting  lines  are  perpendicular  to  each  other. 

65.  From  one  point  in  a  straight  line  only  one  perpendicular 
to  that  line  can  be  drawn  in  the  same  plane ;  or  if  more  be 
drawn,  they  must  coincide. 

Sug.    Consult  Sects.  22,  23,  and  36 ;  also  Theorem  61. 

66.  Hyp.   If  two  adjacent  angles  be  supplementary, 

Con.  their  exterior  sides  form  one  and  the  same  straight 
line. 


14  PLANE  GEOMETET. 

Post.   Let  ABH  and  HBC  be  2  sup.  adj.  A     We  are  to 
prove  that  AB  and  BC  form  one  and  the  same  st.  line. 


A.  B  C 

Dem.  BC  is,  by  Hyp.,  a  st.  line,  therefore  if  it  be  extended 
in  either  direction,  it  will  still  be  a  st.  line.  If  now  we  can 
prove  that  the  extension  of  BG  must  coincide  with  AB,  then 
AB  and  BC  must  form  one  line. 

There  are  only  two  possible  relations,  as  regards  position, 
between  AB  and  the  extension  of  BC-,  i.e.  they  either  coincide 
or  they  do  not.  Now,  if  we  can  prove  the  impossibility  of  all 
the  possible  relations  except  one,  whether  two  or  more,  knowing 
that  one  of  them  must  be  true,  that  is  equivalent  to  a  direct 
proof  of  the  existence  of  that  one  relation,  and  is  just  as  con- 
clusive. 

For  example,  we  know  that  of  the  two  quantities  x  and  y, 
one  of  three  relations  must  be  true;  viz. :  (1)  x  >  y,  (2)  x  =  y, 
(3)  x<y.  There  is  no  other  possible  relation  in  respect  to 
size ;  hence,  if  we  can  prove  the  impossibility  of  the  first  two 
relations  in  any  given  case,  then  that  is  conclusive  that  the 
third  is  true,  and  that  consequently  x  <  y. 

Let  us  apply  this  method  to  the  demonstration  of  this 
theorem. 

It  is  evident  that  AB  is  either  the  extension  of  BC  or  it  is 
not.  Let  us  suppose  that  the  latter  relation  is  true  and  then 
trace  the  consequences.  If  AB  is  not  the  extension  of  BC, 
then  some  other  line,  as  BD,  must  be.  Now,  if  BD  is  the 
extension  of  BC,  CBD  must  be  a  st.  line,  whether  it  seems  to 
the  eye  to  be  so  or  not. 

.-.  £  HBC  +  £  HBD  =  2  rt.  A,  (Sect.  32.) 

but  Z. HBC  +  /.HBA  =  2rt.A-,          (By  Hyp.) 


PLANE  GEOMETRY.  15 


.-,  Z  EEC  +  Z  #££  =  Z  /ZBC  +  Z  JIBJL    (Axiom  III.) 
=  Z#B(7j  (Iden.) 

=  ZHBA.  (Axiom  VII.) 

Let  us  examine  this  last  equation.  Can  the  equation  be  a 
true  one  ?  Can  the  Z.HBD  be  equal  to  the  ZHBA  ?  Evidently 
not,  for  the  former  is  a  part  of  the  latter  (Axiom  I.).  We 
have,  therefore,  arrived  at  an  impassible  result.  -It  is  plain, 
therefore,  that  there  must  be  error  somewhere  ;  either  in  the 
argument  or  in  the  supposition  upon  which  the  argument  is 
based.  A  careful  and  critical  review  of  the  argument  dis- 
closes no  flaw,  no  error,  but,  on  the  contrary,  it  is  in  perfect 
accord  with  previously  established  principles  j  consequently 
the  conclusion  is  inevitable  that  the  error  is  in  our  supposition. 
If,  then,  our  supposition  that  BA  is  not  the  extension  of  BG 
is  erroneous,  the  only  other  possible  relation  must  be  true; 
viz.  that  BA  rs  the  extension  of  BC,  and  therefore  they  must 
form  one  and  the  same  straight  line.  Q.E.D. 

67.  If  two  vertical  angles  be  bisected,  the  bisecting  lines 
form  one  and  the  same  line. 

Sug.    Use  Theorems  64  and  65. 

68.  If  two  pairs  of  vertical  angles  be  bisected,  the  bisecting 
lines  are  perpendicular  to  each  other. 

Sug.   Consult  Theorems  64  and  67. 

69.  The  line  which  bisects  one  of  two  vertical  angles  will, 
if  extended,  bisect  the  other. 


ANGLE  MEASUREMENT. 

70.  The  most  common  unit  for  measuring  the  magnitude 
of  angles  is  the  ninetieth  part  of  a  right  angle,  called  a  degree. 
The  degree  is  subdivided  into  sixty  equal  parts  called  minutes, 
and  the  latter  also  into  sixty  equal  parts  called  seconds.  By 


16  P^ANE  GEOMETRY. 

this  means  small  fractions  of  a  degree  may  be  expressed  in 
minutes  and  seconds.  These  units  are  indicated  by  the  fol- 
lowing symbols ;  viz. : 

0  for  degrees,  '  for  minutes,  and  "  for  seconds  ; 
e.g.  75°  20'  34". 

This  is  called  the  sexagesimal  method.  Another  method, 
called  the  centesimal)  has  been  proposed  in  France,  having 
obvious  advantages.  But  owing  to  the  fact  that  all  mathe- 
matical tables  and  instruments  had  been  arranged  and  con- 
structed with  reference  to  the  sexagesimal  method,  it  has  not 
yet  come  into  extensive  practical  use. 

In  this  method  the  right  angle  is  divided  into  one  hundred 
equal  parts  called  grades,  the  grade  into  the  same  number  of 
equal  parts  called  minutes,  and  the  latter  into  one  hundred  equal 
parts  called  seconds.  These  are  designated  by  symbols  as 
follows ;  viz. : 

45*  15'  28*. 

How  many  degrees  in  a  right  angle  ? 

How  many  grades  in  a  right  angle  ? 

What  is  the  complement  of  a  right  angle  ? 

What  is  the  supplement  of  a  right  angle  ? 

How  many  degrees  in  all  the  angles  formed  at  one  point  on 
one  side  of  a  straight  line  ?  How  many  grades  ? 

How  many  grades  in  all  the  angles  formed  by  any  number 
of  straight  lines  meeting  at  a  common  point  ?  How  many 
degrees  ? 

What  is  the  complement  of  45°;  30°;  1°;  89°  59'  59"  5 
45*;  30*;  1*;  89*59'59n? 

What  is  the  supplement  of  each  of  the  above  angles  ? 

How  many  degrees  in  an  angle  that  is  double  its  comple- 
ment ?  How  many  grades  ? 

How  many  degrees  in  an  angle  that  is  four  times  its  sup- 
plement ?  How  many  grades  ? 


PLANE  GEOMET&Y.  17 

What  angle,  in  degrees  and  grades,  do  the  hour  and  minute 
hands  of  a  clock  form  at  2  o'clock?  At  3  o'clock?  At 
5  o'clock? 

If  one  of  the  angles,  formed  by  two  straight  lines  crossing 
each  other,  be  120°,  find  the  values  of  each  of  the  other  three 
angles  in  degrees. 

If  two  complementary  adjacent  angles  be  bisected,  what 
is  the  value  of  the  angle  formed  by  the  bisectors  in  degrees  ? 
In  grades  ? 

What  is  the  complement  of  37°  44'  51"? 

What  is  the  supplement  of  104°  33'  21"  ? 

What  is  the  complement  of  41«  28'  32"? 

What  is  the  supplement  of  125^  76'  84V? 

An  angle  of  60g  is  how  many  degrees  ? 

An  angle  of  99°  is  how  many  grades  ? 

One-half  a  right  angle  is  what  part  of  three  right  angles? 
Of  two  right  angles  ? 

One-fifth  of  two  right  angles  is  what  part  of  one  right  angle  ? 
Of  four  right  angles  ? 

How  many  degrees  in  an  angle  that  is  one-third  its  comple- 
ment ?  How  many  grades  ? 

If  one  of  two  complementary  angles  is  acute,  what  kind  of 
an  angle  must  the  other  be  ? 

If  one  of  two  supplementary  angles  is  acute,  what  kind 
must  the  other  one  be  ? 

71.  A  perpendicular  is  the  shortest 
line  that  can  be  drawn  from  a  point 
to  a  straight  line. 

(The  analysis  of  this  theorem  has    A.        B 
been  given  in  the  Introduction.)  \ 

Post.    Let  AD  be  any  st.  line,  and  \ 

HC  a  line  drawn  from  pt.  H  _L  to 
AD. 


K 


Also  let  HB  be  any  other  line  that  can  be  drawn  from  pt.  H 
to  AD. 


18  PLANE  GEOMETEY. 

We  are  to  prove  that  HG  <  HB. 

Cons.  Extend  HG  from  pt.  G,  making  GK '  =  CH,  and  join 
BK. 

Dem.  Conceive  the  figure  BHG  to  be  revolved  on  BO  as  an 
axis  until  it  conies  into  the  same  plane  with  figure  BGK. 

Then  the  line  CH  will  fall  upon  GK. 

(Sect.  36  and  Theorem  61.) 

Pt.  H  will  fall  upon  pt.  K.  Why  ? 

Line  BH  will  coincide  with  line  BK.  Why  ? 

What  is  the  relation,  then,  between  BH  and  BK? 

KG+GH<BH  +  BK,  Why? 

or                            GH+GH<BH  +  BH.  How  obtained? 

Whence                       2CH<2  BH,  Why  ? 

or                                         GH<BH.  Why? 

But  BH  represents  any  line  that  can  be  drawn  from  pt.  H 
to  line  AD  other  than  the  perpendicular  CH.  Hence,  from 
our  last  inequality,  the  perpendicular  is  in  every  case  the 
shorter.  Q.E.D. 

72.   The  converse  of  the  previous  theorem ;  viz. : 

Hyp.  If  several  lines  be  drawn  from  a  point  to  a  straight 
line,  one  of  which  is  the  shortest  that  can  be  drawn, 

Con.  the  latter  only  will  be  perpendicular  to  the  said 
straight  line. 

Sug.  Let  AB  be  the  shortest  line  that  can  be  drawn  from 
pt.  A  to  line  CD,  and  from  A  draw  a  perpendicular  to  CD. 

By  Theorem  15,  what  must  this  _L  be  ? 

How  must  it  and  line  AB  be  situated  relatively,  then  ? 

Remarks,  (a)  The  perpendicular  distance  from  a  point  to 
a  line  is  usually  termed  simply  the  distance. 

(b)  When  a  _L  is  drawn  to  another  line,  this  other  line  is 
called  the  base  of  the  _L ;  and  the  pt.  where  the  J_  meets  the 
base  is  called  the  foot  of  the  _L. 


PLANE  GEOMETRY.  19 

73.  If  two  oblique  lines  be  drawn  from  the  same  point  in 
a  perpendicular  to  its  base,  so  as  to  cut  off  equal  distances 
from  the  foot  of  the  perpendicular, 

I.  the  two  oblique  lines  will  be  equal ; 

II.  the  two  angles  which  the  oblique  lines  make  with  the 
perpendicular  are  equal ; 

III.  the  two  angles  formed  by  the  two  oblique  lines  and 
the  base  are  equal. 

Sug.   Use  method  similar  to  that  in  71. 

74.  If  a  perpendicular  be  drawn  to  a  line  at  its  middle  point, 
and  any  point  in  the  perpendicular  be  selected  at  random,  this 
point  is  the  same  distance  from  one  extremity  of  the  given 
line  as  from  the  other. 

75.  If  a  perpendicular  bisect  a  given  line,  and  any  point  out- 
side the  perpendicular  be  selected  at  random,  the  distances  of 
this  point  from  the  extremities  of  the  given  line  are  unequal. 

76.  If  two  lines  be  drawn  from  a  point  to  the  extremities 
of  a  given  line,  and  two  other  lines  be  drawn  from  a  point 
within  the  first  two  to  the  same  extremities,  the  sum  of  the 
first  two  lines  will  exceed  the  sum  of  the  other  two. 

Sug.  Extend  one  of  the  second  pair  until  it  meets  one  of 
the  first  pair. 

77.  If  from  the  same  point  in  a  perpendicular  two  oblique 
lines  be  drawn  to  the  base  so  as  to  cut  off  unequal  distances 
from  the  foot  of  the  perpendicular,  the  one  cutting  the  base 
the  farther  from  the  foot  will  be  the  longer. 

Sug.  In  constructing  diagram  for  this  Dem.  be  sure  that 
both  oblique  lines  are  not  on  the  same  side  of  the  J_.  Consult 
73  and  76. 

78.  Converse  of  73,  I. 

Sug.  Use  method  similar  to  that  illustrated  in  66 ;  i.e.  desig- 
nating the  distances  cut  off  by  x  and  y,  then  x  >  y,  or  x  <  y, 
or  x  =  y.  Prove  that  the  first  two  relations  are  impossible. 


20  PLANE  GEOMETRY. 

79.  Converse  of  77. 

JSng.   Use  method  similar  to  that  in  78. 

80.  Only  two  equal  straight  lines  can  be  drawn  from  the 
same  point  in  a  perpendicular  to  its  base,  or  from  a  point  to  a 
straight  line. 

81.  From  a  point  without  a  straight  line  only  one  perpen- 
dicular to  that  line  can  be  drawn  j  or  if  more  be  drawn,  they 
must  coincide. 

82.  If  two  points,  each  of  which  is  equally  distant  from  the 
extremities  of  a  given  line,  be  joined  by  a  line,  this  line  or  its 
extension  will  bisect  the  given  line  and  be  perpendicular  to  it. 

83.  If  two  lines  in  the  same  plane  be  perpendicular  to  the 
same  straight  line,  they  are  parallel. 

Sug.  They  must  either  be  parallel  or  meet.      Show  the 
impossibility  of  the  latter  condition. 

84.  If  a  line  is  perpendicular  to  one  of  two  parallel  lines, 
it  will  be  perpendicular  to  the  other  also. 

85.  If  two  straight  lines  be  parallel  to  a  third,  they  are 
parallel  to  each  other. 

Sug.   Construct  a  perpendicular  to  one,  and  then  consult  83. 

TKANSVEKSALS. 

86.  When  two  straight  lines  in  the  same  plane  are  crossed 

by  another,  the  latter  is  called 
a  transversal. 

In  the  figure,  which  line  is 
the  transversal  ? 

How  many  angles  are  formed? 

Certain  ones  of  the  above 
angles  are  termed  interior  an- 
gles. Name  them. 

Why  are  they  called  interior 
angles  ? 


PLANE  GEOMETRY.  21 

What  name  would  you  give  to  the  others  to  distinguish 
them  from  those  already  named  ?  Why  ? 

How  many  pairs  of  vertical  angles  ?     Name  them. 

How  many  pairs  of  adjacent  angles  ?     Name  them. 

Two  angles  situated  on  opposite  sides  of  the  transversal, 
either  both  interior  or  both  exterior,  and  not  adjacent,  are 
called  alternate  angles. 

How  many  pairs  of  alternate  interior  angles  ?    Name  them. 

How  many  pairs  of  alternate  exterior  angles  ?    Name  them. 

Two  non-adjacent,  non-vertical  angles,  of  which  one  is  inte- 
rior and  the  other  exterior,  and  both  on  the  same  side  of 
transversal,  are  called  technically  exterior  interior  angles. 

How  many  pairs  of  exterior  interior  angles  ?    Name  them. 

87.   Hyp.  If  two  parallel  lines  be  crossed  by  a  transversal, 

Con.  the  alternate  interior 
angles  are  equal. 

Post.  Let,  etc.  (The  pupil 
should  give  it  with  precision.) 

We  are  to  prove  that  any 
pair  of  Alt.  Int.  angles,  selected 
at  random,  are  equal ;  e.g. : 

that  ZKBC  =  ZBCN. 

\  T»  D 

Cons.  Through  Q,  the  mid- 
dle point  of  BC,  construct  R  W 
perpendicular  to  one  of  the  two  parallels,  as  HK. 

Dem.   What  is  the  position  of  R  W  relative  to  NP  ?    Why  ? 

Conceive  the  entire  figure  AHBQRK  to  revolve  about  Q  as 
a  pivot,  keeping  always  in  the  same  plane,  until  the  line  QA 
coincides  with  QD. 

What  relation  in  regard  to  magnitude  exists  between  the 
angles  BQR  and  WQC  ?  Why  ? 

When  QA  has  reached  the  position  of  QD,  where  must  the 
point  B  be  ?  Why  ? 

What  position  must  the  line  QR  take  ?    Why  ? 


• 

OF  THE 

UNIVER 


22  PLANE  GEOMETRY. 

What  must  be  the  relative  position,  then,  of  the  lines  CW 
and  BE?  Why?  (See  81.) 

What  must  be  the  relative  positions,  then,  of  the  points 
R  and  W?  Why  ? 

What  is  true,  then,  of  the  sides  and  vertices  of  the  two 
angles  KBC  and  BCN? 

What  must  be  the  relation,  then,  in  respect  to  magnitude, 
between  these  two  angles  ? 

Hence,  etc.  Q.E.D. 

88.  If  two  parallel  lines  be  crossed  by  a  transversal,  prove 

I.  that  the  exterior  interior  angles  are  equal ;  (Sug.  Consult 
87  and  56.) 

II.  that  the  alternate  exterior  angles  are  equal ; 

III.  the  two  interior  angles  on  the  same  side  of  the  trans- 
versal are  supplementary ; 

IV.  that  the  two  exterior  angles  on  the  same  side  of  the 
transversal  are  supplementary. 

89.  Hyp.   If  two  lines  in  the  same  plane  be  crossed  by  a 
transversal  so  as  to  make  the  exterior  interior  angles  equal, 

Con.  these  two  lines  will  be  parallel. 


Post.  Let  AC  and  JVP  be  two  lines  in  the  same  plane 
crossed  by  the  transversal  DH,  making  the  angles  DBA  and 
BKN  equal. 

We  are  to  prove  that  AC  and  NP  are  parallel. 

Dem.  It  is  evident  that  they  must  occupy  one  of  the  two 
relative  positions ;  viz.  parallel  or  non-parallel. 


PLANE  GEOMETRY.  23 

Let  us  adopt  the  supposition  that  they  are  non-parallel. 
Then  a  line  may  be  drawn  through  point  K,  parallel  to  AC. 
Let  the  dotted  line  ST  represent  this  line. 

Then  Z  DBA  =  Z  BKS.  (Theorem  88  I. ) 

But  Z  DBA  =  Z  BKN.  (By  Hyp.) 

.-.  Z  BKS  =  Z  BKN.  (Axiom  III.) 

The  pupil  should  finish  the  argument. 

Bug.  Consult  the  argument  in  the  demonstration  of  Theorem 
66. 

90.   Hyp.   If  two  lines  in  the  same  plane  be  crossed  by  a 
transversal  so  as  to  make  the  alternate  interior  angles  equal, 
Con.   these  two  lines  will  be  parallel. 

-A 


Post.   Let  AB  and  CD  be  two  lines  in  the  same  plane 
crossed  by  the  transversal  HQ,  making  Z  CNK~  Z  NKB. 
We  are  to  prove  that  AB  and  CD  are  parallel. 

Dem.                      Z.CNK=^QND.  Why? 

Z  CNK  =  Z  NKB.  Why  ? 

.-.  Z  QND  =  Z  #KB.  Why  ? 

.-.  CD  and  .45  are  parallel.  (Theorem  89.) 

91.   Converse  of  38,  II.,  III.,  and  IV. 

91  (a).   If  two  lines  be  crossed  by  a  transversal  making 

I.  the  alternate  interior  or  exterior  angles  unequal,  or 

II.  the  exterior  interior  angles  unequal,  or 

III.  the  two  interior,  or  two  exterior,  angles,  on  the  same  side 
of  the  transversal,  together  less  than  a  straight  angle,  the  two 


24  PLANE  GEOMETRY. 

lines  will  meet  if  sufficiently  extended,  and  on  that  side  of 
the  transversal  on  which  the  sum  of  the  angles  is  less  than  a 
straight  angle. 

92.  If  two  parallel  lines  be  crossed  by  a  transversal,  the 
bisectors  of  the  alternate  interior,  or  alternate  exterior,  angles 
are  parallel. 

93.  If  two  parallel  lines  be  crossed  by  a  transversal,  the 
bisectors  of  the  two  interior,  or  the  two  exterior  angles,  on  the 
same  side  of  the  transversal,  are  perpendicular  to  each  other. 

Sug.   Consult  64,  92,  and  84. 

94.  If  two  angles  have  their  sides  parallel,  two  and  two, 
they  are  either  equal  or  supplementary. 

95.  If  two  angles  have  their  sides  respectively  perpendicular 
to  each  other,  these  angles  are  either  equal  or  supplementary. 

Which  of  the  above  results  will  be  true  if  the  angles  are 
both  acute  ?  If  both  obtuse  ?  If  one  is  acute  and  the  other 
obtuse  ?  If  both  are  right  angles  ? 

TKIANGLES, 

96.  A  triangle  is  a  plane  figure  bounded  by  three  straight 
lines,  and  consequently  has  three  angles.     It  is  sometimes 
called  a  trigon. 

Triangles  are  named, 

I.  with  reference  to  the  character  of  their  angles ; 

II.  with  reference  to  the  relations  between  their  sides. 

I.  If  a  triangle  has  all  its  angles  acute,  it  is  called  an  acute 
triangle. 

If  it  has  one  obtuse  angle,  it  is  called  an  obtuse  triangle. 
If  it  has  one  right  angle,  it  is  called  a  right  triangle. 

II.  If  all  its  sides  are  equal,  it  is  called  an  equilateral  triangle. 
If  two  of  its  sides  are  equal,  it  is  called  an  isosceles  triangle. 
If  no  two  of  its  sides  are  equal,  it  is  called  a  scalene  triangle. 


PLANE  GEOMETRY.  25 

The  term  isosceles  is  used  to  designate  the  fact  that  two 
sides  are  equal  without  any  reference  to  the  third  side,  even 
though  it  may  then  be  known,  or  afterwards  ascertained,  that 
the  sides  are  all  equal. 

Construct  triangles  to  illustrate  each  of  the  above  varieties. 

Construct  three  triangles  that  shall  illustrate  all  the  six 
varieties. 

The  sum  of  the  three  sides  of  a  triangle  is  called  the  perim- 
eter. 

In  a  right  triangle,  the  side  opposite  the  right  angle  is  called 
the  hypothenuse,  and  the  other  two  sides  the  legs. 

The  side  on  which  a  triangle  is  supposed  to  rest  is  called  the 
base.  In  general,  any  side  of  a  triangle  may  be  considered 
the  base ;  but  in  an  isosceles  triangle  the  third  side,  and  in  a 
right  triangle  one  of  the  legs,  is  generally  the  base. 

The  angle  opposite  the  base  is  called  its  vertical  angle ;  and 
its  vertex,  the  vertex  of  the  triangle. 

The  perpendicular  distance  from  vertex  to  base,  or  base 
extended,  is  called  the  altitude  of  a  triangle. 

The  line  drawn  from  the  vertex  of  a  triangle  to  the  middle 
point  of  the  base  is  called  a  median. 

97.  Any  side  of  a  triangle  is  less  than  the  sum  of  the  other 
two,  and  greater  than  their  difference. 

98.  The  sum  of  the  three  angles  of  a  triangle  is  equal  to 
two  right  angles. 

Sug.   Consult  Theorems  58  and  87. 

99.  If  one  side  of  a  triangle  be  extended,  the  exterior  angle 
thus  formed  is   equal  to  the  sum  of  the  two  interior  non- 
adjacent  angles. 

100.  If  two  angles  of  a  triangle  be  known,  how  can  the  third 
be  found  ? 

Two  angles  of  a  triangle  are  34°  28' 42"  and  29°  44' 56", 
respectively ;  find  the  value  of  the  third  angle. 


26  PLANE  GEOMETRY. 

Two  angles  of  a  triangle  are  64*75W  and  86g45r75M, 
respectively ;  find  value  of  the  third  angle. 

If  one  angle  of  a  triangle  be  a  right  angle,  what  relation 
exists  between  the  other  two  ? 

How  many  right  angles  in  a  triangle  ?  How  many  obtuse 
angles  ? 

If  the  three  angles  of  a  triangle  are  all  equal,  what  is  the 
value  of  each  angle  in  terms  of  a  right  angle  ?  In  degrees  ? 
In  grades  ? 

If  two  angles  of  one  triangle  be  equal  respectively  to  two 
angles  of  another,  what  relation  exists  between  their  third 
angles  ?  Prove. 

If  one  acute  angle  of  a  right  triangle  is  27°  38'  50",  what  is 
the  value  of  the  other  acute  angle  ? 

Find  values  of  the  three  angles  of  a  triangle,  if  the  second 
is  three  times  the  first  and  the  third  is  two  times  the  second. 

In  a  right  triangle  one  of  the  acute  angles  is  17°  40'  greater 
than  the  other ;  find  the  values  of  both  acute  angles. 

The  sum  of  two  angles  is  30g,  and  their  difference  is  9°; 
find  each  angle. 

101.  Hyp.  If  two  triangles  have  two  sides  and  their  in- 
cluded angle,  of  one,  equal  respectively  to  two  sides  and  their 
included  angle,  of  the  other, 

Con.  the  two  triangles  will  be  equal  in  all  respects. 


Post.  Let  ABC  and  DHK  be  two  triangles  having  side 
AB=DH,  AC=DK,  and  their  included  angles  A  and  D 
equal. 

We  are  to  prove  that  the  triangle  ABC  equals  the  triangle 


PLANE  GEOMETRY.  27 

DHK\  i.e.  that  all  their  remaining  corresponding  parts  are 
respectively  equal. 

Dem.  Conceive  the  triangle  ABC  to  be  so  applied  to  the 
triangle  DHK  that  side  AB  shall  fall  upon  DH,  the  point  A 
falling  on  point  D. 

Where  must  the  pt.  B  fall  ?     Why  ? 

Where  must  the  line  AC  fall  ?     Why  ?  (Sect.  36.) 

Where  must  pt.  C  fall  ?     Why  ? 

What,  then,  must  be  the  relative  position  of  BC  and  HK? 
Why? 

Hence,  the  two  triangles  are  coincident,  and  are  therefore 
equal.  (Sect.  34.) 

Q.E.D. 

102.  If  two  triangles  have  two  angles  and  the  side  con- 
necting their  vertices,  of  one,  equal  respectively  to  two  angles 
and   side  connecting  their  vertices,   of  the  other,   the  two 
triangles  will  be  equal  in  every  respect. 

Sug.    Use  method  similar  to  the  preceding. 

103.  If  the  three  sides  of  one  triangle  are  equal  respectively 
to  the  three  sides  of  another,  the  two  triangles  are  equal  in 
every  respect. 

Sug.  Apply  the  triangles  as  indicated  below,  then  consult 
82,  73,  IL,  and  101. 


\ 

\ 

~JC 

H 

104.   If  two  right  triangles  have  their  legs  respectively 
equal,  they  are  equal  in  all  respects. 


28  PLANE   GEOMETRY. 

105.  If  two  right  triangles  have  a  leg  and  an  acute  angle 
of  one  equal  respectively  to  a  leg  and  an  acute  angle  of  the 
other,  they  are  equal  in  all  respects. 

106.  If  two  right  triangles  have  the  hypothenuse  and  an 
acute  angle  of   one   equal   respectively  to  the   hypothenuse 
and   an    acute    angle   of    the   other,   they   are   equal  in  all 
respects. 

Query.  Would  two  right  triangles  be  equal  if  they  had 
d  side  and  an  acute  angle  of  one  equal  to  a  side  and  an  acute 
angle  of  the  other  ? 

107.  If  two  right  triangles  have  the  hypothenuse  and  a  leg 
of  one  equal  respectively  to  the  hypothenuse  and  a  leg  of  the 
other,  they  are  equal  in  all  respects. 

Sug.  Apply  one  triangle  to  the  other  so  that  the  equal  legs 
shall  coincide  ;  then  consult  78  and  103. 

Remark.  The  corresponding  sides  and  angles  of  two  tri- 
angles are  those  that  are  similarly  situated,  and  are  called 
homologous.  If  the  triangles  are  equal,  the  homologous  sides 
are  those  that  are  opposite  the  equal  angles ;  and  conversely, 
the  homologous  angles  are  those  that  are  opposite  the  equal 
sides. 

108.  Hyp.   If  a  triangle  have  two  of  its  sides  equal, 
Con.   the  angles  opposite  those  sides  are  also  equal. 

Post.  Let  ACD  be  a  triangle 
having  the  sides  AD  and  CD 
equal. 

We  are  to  prove  that 


Cons.   Draw  DB  J_  to  AC. 

Dem.  The  two  triangles  CBD 
and  ABD  are  rt.  A.     Why  ? 
Name  the  hypothenuse  of  each. 
What  line  is  a  leg  of  both  triangles  ? 


PLANE  GEOMETRY.  29 

.-.  The  two  A  are  equal.  (Theorem  107.) 

.*.  Z  G  =  Z  A.  (Being  homologous  angles  of  equal  A.) 

Q.E.D. 

Query.  In  the  above  demonstration  how  else  could  the  line 
BD  have  been  drawn  ?  Give  the  corresponding  variation  in 
demonstration. 

The  pupil  should  also  understand  now,  at  the  outset,  that 
only  one  condition  can  be  imposed,  in  drawing  an  auxiliary 
line,  which  fixes  its  direction.  For  instance,  he  cannot  say,  in 
the  previous  construction,  "Draw  DB  perpendicular  to  CA 
through  its  middle  point,"  since  either  condition  gives  the 
line  a  definite  direction;  and,  previous  to  demonstration,  he 
does  not  know  but  that  one  condition  will  conflict  with  the 
other. 

109.  Converse  of  108. 

Sug.   Employ  method  similar  to  that  in  50. 

110.  If  a  triangle  be  equilateral,  it  will  also  be  equiangular. 
Sug.  Consult  108. 

.  111.   Converse  of  110. 
Sug.   Consult  109. 

112.  If  a  perpendicular  bisect  the  base  of    an  isosceles 
triangle,  it  will  pass  through  the  vertex  and  bisect  the  vertical 
angle. 

113.  Converse  of  112. 

114.  If  a  line  be  drawn  from  the  vertex  of  an  isosceles 
triangle  to  the  middle  point  of  the  base,  this  line  will  bisect 
the  vertical  angle  and  be  perpendicular  to  the  base. 

115.  If  one  of  the  equal  sides  of  an  isosceles  triangle  be 
extended  at  the  vertex,  and  the  exterior  angle  thus  formed 
be  bisected,  the  bisecting  line  will  be  parallel  to  the  base. 

116.  Converse  of  115. 


30  PLANE  GEOMETRY. 

117.  If  a  line  bisect  an  angle,  any  point  selected  at  random 
in  this  bisector  will  be  equally  distant  from  the  sides  of  the 
angle. 

JSug.   Consult  72,  Remark  (a),  and  106. 

118.  If  a  line  bisect  an  angle,  any  point  selected  at  random 
outside  this  bisector  is  unequally  distant  from  the  sides  of 
the  angle. 


Sug.  If  K  be  the  point,  then  what  is  the  relation  between 
HS  and  HB  ?     Why  ? 

What  is  the  relation  between  BK  and  KH  +  HS?    Why  ? 

What  is  the  relation  of  KS  to  KH  +  HS?     Why  ? 

What  is  its  relation,  then,  to  BK? 

What  is  the  relation  between  KS  and  KP  ? 

What,  then,  must  be  the  relation  between  KP  and  BK? 

119.  If  a  point  be  equally  distant  from  the  sides  of  an 
angle,  the  line   joining  it  with  the   vertex  will  bisect  the 
angle. 

120.  If  the  angles  at  the  base  of  an  isosceles  triangle  be 
bisected,  the  bisectors  will  form,  with  the  base,  an  isosceles 
triangle. 

121.  If  the  angles  at  the  base  of  an  isosceles  triangle  be 
double  the  vertical  angle,  a  line  bisecting  either  of  the  former 
will  divide  the  triangle  into  two  isosceles  triangles. 


PLANE  GEOMETRY. 


31 


122.  If  two  angles  of  a  triangle  be  unequal,  the  side  opposite 
the  greater  of  these  two  angles  is  longer  than  the  side  opposite 
the  lesser. 

Post.  Let  ABC  be  any  tri- 
angle with  Z  B  >  Z  A. 

We  are  to  prove  '  that  the 
side  AC,  opposite  the  greater 
angle  B,  is  longer  than  the 
side  CB,  opposite  the  lesser 
angle  A. 

Cons.  Let  BD  be  drawn  so 
as  to  cut  off  a  portion  of  the  larger  angle  B,  making  it  equal 
to  angle  A,  so  that  they  will  both  be  in  the  same  triangle 
as  DAB. 

What  relation  between  AD  and  DB  ?    Why  ? 

What  relation  between  BC  and  CD-}-DB?    Why  ? 

(The  pupil  should  finish  it  without  difficulty.) 

123.  Converse  of  122. 

Sug.  Three  possible  relations  between  the  two  angles. 
Prove  the  impossibility  of  two  of  them  ;  or,  a  method  similar 
to  that  in  122  may  be  employed. 

123  (a).  If  two  triangles  have  the  two  sides  of  one  equal 
respectively  to  two  sides  of  the  other,  but  the  included  angles 
unequal,  then  the  third  side  of  that  triangle  having  the  greater 
included  angle  is  longer  than  the  third  side  of  the  other. 


Sug.  Apply  the  two  triangles  so  that  two  of  the  equal  sides 
shall  coincide,  as  is  represented  in  the  above  diagram. 


THE 


32  PLANE  GEOMETRY. 

Will  the  other  pair  of  equal  sides  coincide  ?    Why  ? 

Join  the  other  two  vertices. 

What  kind  of  a  triangle  is  HKB'? 

What  relation,  then,  between  angles  HKB1  and  HB'K? 

What  must  be  the  relation,  therefore,  between  the  angles 
DKB'  and  C'B'K? 

Consult  Theorem  122. 

Or,  instead  of  joining  B1  and  K,  the  angle  KA'B'  may  be 
bisected,  and  the  point  where  this  bisector  cuts  DK  joined 
with  Bf.  Then  consult  101  and  97. 

124.  Converse  of  123. 

Sug.  There  are  only  three  possible  relations  between  those 
angles.  Prove  the  impossibility  of  two  of  them. 

ADVANCE  THEOREMS. 

125.  The  bisectors  of  the  three  angles  of  a  triangle  meet  in 
one  point. 

Sug.  From  the  point  of  intersection  of  two  of  the  bisectors 
draw  a  line  to  the  other  vertex,  and  also  perpendiculars  to  the 
three  sides.  Prove  the  former  a  bisector  by  means  of  equality 
of  triangles. 

126.  The  perpendiculars  which  bisect  the  three  sides  of  a 
triangle  meet  in  a  point. 

Sug.  From  the  point  of  intersection  of  two  of  the  perpen- 
diculars draw  a  line  to  the  middle  point  of  the  other  side. 
Prove  this  line  perpendicular,  by  consulting  74  and  103. 

127.  The  perpendiculars  from  the  three  vertices  of  a  triangle 
to  the  opposite  sides  meet  in  one  point, 

Sug.  Through  each  vertex  draw  a  line  parallel  to  the 
opposite  side.  Then  consult  102  and  127. 

128.  If  two  angles  of  an  equilateral  triangle  be  bisected, 
and  lines  be  drawn  through  the  point  of  intersection  parallel 
to  the  sides,  the  sides  will  be  trisected. 


PLANE  GEOMETRY.  33 

QUADEILATEKALS, 

129.  If  two  lines  in  the  same  plane  be  crossed  by  two  trans- 
versals, a  figure  of  four  sides  may  be  formed,  which  is  called 
a  quadrilateral.  Hence  a  quadrilateral  may  be  defined  as  a 
plane  figure  bounded  by  four  straight  lines.  A  quadrilateral 
is  also  called  a  tetragon. 

If  each  of  the  two  pairs  of  lines  is  parallel)  the  quadrilateral 
thus  formed  is  called  a  parallelogram. 

Define,  then,  a  parallelogram. 

If  a  parallelogram  have  all  its  sides  equal,  it  is  called  a 
rhombus. 

If  its  angles  are  right  angles,  it  is  called  a  rectangle. 

If  it  is  both  a  rhombus  and  a  rectangle,  it  is  called  a 
square. 

Give  all  the  names  applicable  to  a  square. 

If  a  quadrilateral  have  only  two  of  its  sides  parallel,  it  is 
called  a  trapezoid. 

If  no  two  of  its  sides  are  parallel,  it  is  called  a  trapezium. 

A  parallelogram  whose  angles  are  oblique  and  adjacent  sides 
unequal  is  sometimes  called  a  rhomboid. 

A  rectangle  whose  adjacent  sides  are  unequal  is  sometimes 
called  an  oblong. 

The  line  which  joins  two  opposite  vertices  of  a  quadrilateral 
is  called  its  diagonal. 

The  side  upon  which  a  parallelogram  is  conceived  to  rest, 
and  the  side  opposite  the  latter,  are  termed,  respectively,  the 
lower  and  upper  bases. 

The  parallel  sides  of  a  trapezoid  are  always  considered  as 
its  bases,  the  other  two  sides  its  legs,  while  the  line  bisecting 
the  legs  is  called  the  median. 

The  altitude  of  either  a  parallelogram  or  trapezoid  is  the 
perpendicular  distance  between  its  bases. 

The  pupil  should  construct  figures  to  represent  all  the  above- 
mentioned  quantities. 


34  PLANE  GEOMETRY. 

The  sum  of  the  four  sides  of  a  quadrilateral  is  called  its 
perimeter. 

If  the  legs  of  a  trapezoid  are  equal,  it  is  called  an  isosceles 
trapezoid. 

130.  Either  diagonal   divides   the   parallelogram   into   two 
equal  triangles. 

Sug.    Consult  87  and  102. 

Query.   Is  the  converse  of  this  theorem  true  ? 

131.  The  sum  of  the  four  angles  of  a  quadrilateral  equals 
four  right  angles. 

132.  The  opposite  sides  of  a  parallelogram  are  equal. 

132  (a).   Two  parallel  lines  are  everywhere  equally  distant. 

133.  Converse  of  132. 

tSug.  Draw  one  diagonal,  then  consult  45  and  34. 

134.  The  opposite  angles  of  a  parallelogram  are  equal. 

135.  Converse  of  134. 
Sug.   Consult  131  and  91. 

136.  If  two  sides  of  a  quadrilateral  are  equal  and  parallel, 
the  quadrilateral  is  a  parallelogram. 

137.  The  two  diagonals  of  a  parallelogram  bisect  each  other. 

138.  Converse  of  137. 

139.  If  one  angle  of  a  parallelogram  be  a  right  angle,  the 
other  three  angles  are  right  angles,  and  the  parallelogram  is 
therefore  a  rectangle. 

140.  The  diagonals  of  a  rectangle  are  equal. 

141.  Converse  of  140. 

142.  The  diagonals  of  a  rhombus  are  perpendicular  to  each 
other. 

143.  Converse  of  142. 

144.  The  diagonals  of  a  rhombus  bisect  the  angles. 

145.  Converse  of  144. 


PLANE  GEOMETRY.  35 

146.  If  two  parallel  lines  be  crossed  by  a  transversal,  the 
bisectors  of  the  interior  angles  form  a  rectangle. 

Sug.   Consult  88,  III.,  92,  93,  and  64. 

147.  The  lines  which  bisect  the  angles  of  a  rhomboid  form 
a  rectangle. 

148.  If  two  parallelograms  have  two  sides  and  their  in- 
cluded angle  of  one  equal  respectively  to  two  sides  and  their 
included  angle  of  the  other,  the  two  parallelograms  are  equal 
in  every  respect. 

149.  If  a  line  be  drawn  parallel  to  one  side  of  a  triangle, 
and  bisecting  another  side,  this  line  will  bisect  the  third  side 
of  the  triangle,  and  be  equal  to  one-half  the  side  parallel  to  it. 

150.  Converse  of  149. 

151.  The  median  of  a  trapezoid  is  parallel  to  its  bases  and 
equal  to  one-half  their  sum. 

Sug.   Through  one  extremity  of  the  median  draw  a  line 
parallel  to  one  leg,  and  extend  the  shorter  base  to  meet  it. 

ADVANCE  THEOREMS. 

152.  If  from  any  point  in  the  base  of  an  isosceles  triangle 
lines  are  drawn  parallel  to  the  equal  sides,  the  perimeter  of 
the  parallelogram  thus  formed  will  be  equal  to  the  sum  of 
the  two  equal  sides  of  the  triangle. 

153.  If  the  legs  of  a  trapezoid  are  equal,  the  angles  which 
they  make  with  either  base  are  equal. 

154.  Converse  of  153. 

155.  If  the  angles  at  one  base  of  a  trapezoid  are  equal,  the 
angles  at  the  other  base  are  also  equal. 

156.  The  line  which  is  parallel  to  the  bases  of  a  trapezoid 
and  bisects  one  leg  is  a  median. 


36  PLANE  GEOMETRY. 

157.  The  line  joining  the  vertex  of  the  right  angle  to  the 
middle  point  of  the  hypothenuse  in  a  right  triangle  is  equal 
to  one-half  the  hypothenuse. 

158.  The  lines  which  join  the  middle  points  of  the  sides 
of  a  triangle  divide  the  triangle  into  four  equal  triangles. 

159.  The  three  medians  of  a  triangle  meet  in  one  point. 
Sug.   From  the  point  of  intersection  of  the  two  medians 

draw  a  line  to  the  third  vertex.  From  the  middle  point  of 
this  line  draw  another  to  the  point  midway  between  one  of 
the  other  vertices  and  the  point  of  intersection  of  the  two 
medians.  By  extending  the  first  line  and  connecting  certain 
points  a  parallelogram  may  be  formed. 

160.  The  lines  which  join  the  middle  points  of  the  sides 
of  any  quadrilateral  form  a  parallelogram. 

Sug.   Draw  the  diagonals,  then  consult  150. 

161.  The  lines  which  join  the  middle  points  of  the  sides  of 
a  rhombus  form  a  rectangle. 

162.  The  lines  which  join  the  middle  points  of  the  sides  of 
a  square  form  a  square. 

163.  The  lines  which  join  the  middle  points  of  the  sides 
of  a  rectangle  form  a  rhombus. 

164.  The  lines  which  join  the  middle  points  of  the  sides 
of  an  isosceles  trapezoid  form  a  rhombus. 

165.  The  median  and  diagonals  of  a  trapezoid  intersect  at 
the  same  point. 

Sug.    Consult  151  and  149. 

166.  The  diagonals  of  an  isosceles  trapezoid  are  equal. 

167.  Converse  of  166. 

168.  If  a  trapezoid  be  isosceles,  the  opposite  angles  are 
supplementary. 


PLANE  GEOMETRY.  37 

169.  The  line  which  joins  the  middle  points  of  the  diagonals 
of  a  trapezoid  equals  one-half  the  difference  of  the  bases. 

170.  The  two  perpendiculars  from  the  extremities  of  the 
base  to  the  equal  sides  of  an  isosceles  triangle  are  equal. 

171.  The  medians  drawn  to  the  equal  sides  of  an  isosceles 
triangle  are  equal. 

172.  The  bisectors  of  the  angles  at  the  base  of  an  isosceles 
triangle  are  equal. 

173.  The  two  perpendiculars  from  the  middle  point  of  the 
base  of  an  isosceles  triangle  to  the  equal  sides  are  equal. 

174.  State  and  prove  the  converse  of  each  of  170, 171,  172, 
173. 

175.  If  one  of  the  equal  sides  of  an  isosceles  triangle  be 
extended  at  the  vertex,  making  the  extension  equal  to  the 
side,  the  line  joining  the  end  of  the  extension  with  the  nearer 
extremity  of  the  base  is  perpendicular  to  the  base. 

176.  If  one  angle  of  an  isosceles  triangle  be  60°,  the  triangle 
is  equilateral. 

177.  If  from  the  middle  points  of  two  opposite  sides  of  a 
parallelogram  lines  be  drawn  to  the  vertices  of  the  angles 
opposite,  these  lines  will  trisect  the  diagonal  that  joins  the 
other  two  vertices. 

178.  If  the  two  base  angles  of  a  triangle  are  bisected,  and 
through  the  point  of  intersection  of  these  bisectors  a  line  be 
drawn  parallel  to  the  base  and  terminating  in  the  sides,  this 
line  is  equal  to  the  sum  of  the  two  segments  of  the  sides 
between  this  parallel  and  the  base. 

179.  The  bisectors  of  the  vertical  angle  of  a  triangle  and 
the  angles  formed  by  extending  the  sides  below  the  base  meet 
in  a  point  which  is  equally  distant  from  the  base  and  the 
extensions  of  the  sides. 


38  PLANE  GEOMETRY. 

180.  If  one  of  the  acute  angles  of  a  right  triangle  is  double 
the  other,  the  hypothenuse  is  double  the  shorter  leg. 

181.  If  from  any  two  points  selected  at  random  in  the  base 
of  an  isosceles  triangle  perpendiculars  be  drawn  to  the  equal 
sides,  the  sum  of  the  perpendiculars  from  one  point  equals  the 
sum  of  the  perpendiculars  from  the  other  point. 

182.  If  from  any  point  selected  at  random  in  an  equilateral 
triangle  perpendiculars  be  drawn  to  the  sides,  the  sum  of  these 
perpendiculars  is  constant,  and  equal  to  the  altitude  of  the 
triangle. 

CIRCLES. 

183.  A  circle  is  a  portion  of  a  plane  bounded  by  a  curved  line, 
all  points  of  which  are  equally  distant  from  a  point  within 
called  the  centre. 

The  bounding  line  is  called  the  circumference. 

Any  portion  of  the  circumference  is  called  an  arc. 

A  radius  (plural  radii)  is  any  line  from  centre  to  circum- 
ference. 

A  diameter  is  any  line  passing  through  the  centre  and  ter- 
minating both  ways  in  tho  circumference. 

How  do  the  radius  and  diameter  of  the  same  circle  compare 
in  magnitude  ? 

How  do  the  radii  of  a  circle  compare  in  magnitude  ?  The 
diameters  ? 

A  semicircumference  is  one-half  the  circumference. 

A  sector  is  that  part  of  a  circle  included  between  an  arc  and 
the  radii  drawn  to  its  extremities. 

A  quadrant  is  a  sector  which  is  one-fourth  the  circle. 

A  quadrant  arc  is  one-fourth  the  circumference. 

A  chord  is  any  straight  line  whose  extremities  are  in  the 
circumference. 

A  segment  is  that  portion  of  the  circle  included  between  an 
arc  and  the  chord  which  joins  its  extremities. 


PLANE  GEOMETRY.  39 

Every  chord,  therefore,  must  divide  the  circumference  into 
two  arcs  and  the  circle  into  two  segments. 

If  the  arcs  are  unequal,  they  are  designated  as  major  and 
minor  arcs,  and  the  segments  as  major  and  minor  segments. 

The  chord  is  said  to  subtend  the  arc,  and  the  arc  is  said  to 
be  subtended  by  the  chord. 

Whenever  a  chord  and  its  subtended  arc  are  mentioned,  the 
minor  arc  is  meant  unless  it  is  otherwise  specified. 

If  two  circles  have  the  same  centre,  they  are  said  to  be 
concentric. 

A  central  angle  is  an  angle  formed  by  two  radii. 

An  inscribed  angle  is  an  angle  formed  by  two  chords,  with 
its  vertex  in  the  circumference. 

When  would  an  angle  be  said  to  be  inscribed  in  a  segment  ? 

A  tangent  is  a  straight  line  that  touches  the  circumference  of 
a  circle,  but  on  being  extended  does  not  intersect  it ;  i.e.  the 
tangent  and  the  circumference  have  one  point,  and  only  one, 
in  common.  This  point  is  called  the  point  of  contact,  or  point 
of  tangency. 

Two  circumferences  are  tangent  to  each  other  when  they 
have  one  point  in  common  but  do  not  intersect ;  i.e.  when  they 
touch  each  other. 

If  one  of  two  tangent  circumferences  lies  within  the  other, 
they  are  said  to  be  tangent  internally;  if  it  lies  without,  they 
are  said  to  be  tangent  externally. 

A  secant  is  a  straight  line  that  intersects  a  circumference  in 
two  points  lying  partly  within  and  partly  without  the  circle ; 
e.g.  a  chord  extended  in  either  direction  becomes  a  secant. 

The  term  circle  is  also  sometimes  used  to  designate  a  cir- 
cumference. 

Construct  diagrams  illustrating  the  above  magnitudes  and 
their  relations. 

184.   A  diameter  is  greater  than  any  other  chord. 
Sug.   Draw  the  diameter  AD,   and  let  HB  be  any  other 
chord.     Join  CH  and  CB.     Consult  97. 


40  PLANE  GEOMETRY. 

185.   A  diameter  bisects  both  the  circle  and  circumference. 

Sug.  Fold  over  one  part  on  the 
other,  using  the  diameter  as  an  axis 
of  revolution. 

185  (a).   Converse  of  185. 

186.  A  straight  line  cannot  intersect 
the  circumference  of  a  circle  in  more 
than  two  points. 
Sug.    Suppose  it  could  intersect  in  three  points,  and  draw 
radii  to  the  three  points ;  then  consult  80. 

187.  If  two  circles  have  equal  radii,  they  are  equal. 
Bug.   Apply  one  to  the  other. 

188.  If  the  diameters  of  two  circles  are  equal,  the  circles 
are  equal. 

189.  Converse  of  187  and  188. 

190.  If  two  equal  circles  be  concentric,  their  circumferences 
will  coincide. 

191.  If,  in  the  same  or  equal  circles,  two  central  angles  be 
equal,  the  arcs  which  their  sides  intercept  will  also  be  equal. 

Remark.  In  this  and  the  following  theorems  where  the 
expression  "  same  or  equal  circles  "  occurs,  the  demonstration 
will  be  more  satisfactory  if  two  circles  are  used. 

Sug.    Apply  one  circle  to  the  other. 

192.  Converse  of  191. 

193.  If,  in  the  same  or  equal  circles,  two  chords  be  equal, 
the  arcs  which  those  chords  subtend  are  also  equal. 

Sug.  Draw  radii  to  the  extremities  of  the  chords,  then  con- 
sult 103  and  191. 

194.  Converse  of  193. 

Sug.   Draw  radii  as  above.     Consult  192  and  101. 


PLANE  GEOMETRY.  41 

195.  If,  in  the  same  or  equal  circles,  two  central  angles  be 
unequal,  the  arc  intercepted  by  the  sides  of  the  greater  angle 
is  greater  than  the  arc  intercepted  by  the  sides  of  the  lesser 
angle. 

Sug.   Apply  one  circle  to  the  other. 

196.  Converse  of  195. 

197.  If,  in  the  same  or  equal  circles,  two  chords  be  unequal, 
the  arc  subtended  by  the  greater  chord  is  greater  than  that 
subtended  by  the  lesser. 

Sug.   Draw  radii  to  extremities,  then  consult  124  and  195. 

198.  Converse  of  197. 

Sug.   Draw  radii  as  above,  then  consult  196  and  123.     Or, 
three  possible  relations. 

199.  If  a  diameter  (or  radius)  be  perpendicular  to  a  chord, 
it  will  bisect  the  chord,  and  also  the  arcs  into  which  the  chord 
divides  the  circumference. 

Sug.   Draw  radii  to  extremities  of  the  chord,  then  consult 
78  or  107  and  191. 

200.  A  straight  line  that  is  perpendicular  to  a  radius  at  its 
extremity  is  a  tangent  to  the  circle. 


Post.   Let  BHKbe  a  circle,  DB  a  radius,  and  AC  a  straight 
line  perpendicular  to  DB  and  passing  through  the  point  B. 
We  are  to  prove  that  AC  is  a  tangent  to  the  circle  BHK. 


42  PLANE  GEOMETRY. 

Now,  if  we  can  prove  that  every  point  in  AC  except  B  is 
out  vide  the  circumference,  then  AC  must  be  a  tangent.  (See 
Del  of  Tangent.) 

Cons.   From  D  draw  any  other  line  to  AC,  as  DP. 

Then  P,  the  extremity  of  DP,  must  represent  any  point  in 
AC  except  B. 

Now  DP  >  DB.     Why  ? 

But  DB  is  a  radius,  and  if  DP  is  longer  than  a  radius, 
where  must  its  extremity  be  ? 

Hence,  etc. 

201.  If  a  radius  be  drawn  to  the  point  of  contact   of  a 
tangent  to  a  circle,  it  will  be  perpendicular  to  the  tangent. 

Sug.  If  it  can  be  proved  that  the  radius  is  the  shortest 
distance  from  the  centre  to  the  tangent,  then  it  must  be  per- 
pendicular. (See  Theorem  72.)  Use  previous  diagram. 

What  must  be  the  situation  of  all  the  points  in  the  tangent, 
except  the  point  of  contact,  with  reference  to  the  circum- 
ference of  the  circle  ? 

What,  then,  must  be  the  relation  to  the  radius  of  a  line 
drawn  from  the  centre  to  any  point  in  the  tangent  except  the 
point  of  contact  ? 

Hence,  etc. 

202.  If  a  perpendicular  be  erected  to  a  tangent  at  the  point 
of  contact,  this  perpendicular,  if  extended,  will  pass  through 
the  centre  of  the  circle. 

Sug.  Draw  a  radius  to  point  of  contact,  then  consult  201 
and  65. 

203.  If  a  line  be  drawn  from  the  centre  of  a  circle  perpen- 
dicular to  a  tangent,  this  line  will  pass  through  the  point  of 
contact. 

204.  If  a  chord  and  tangent  be  parallel,  the  arcs  which  they 
intercept  are  equal. 

Sug.  Draw  a  radius  to  point  of  contact,  then  consult  201, 
84,  and  199. 


PLANE  GEOMETRY.  43 

205.  If  two  chords  be  parallel,  the  arcs  which  they  inter- 
cept are  equal. 

206.  If  two  tangents  be  parallel,  the  arcs  which  they  in- 
tercept are  equal. 

207.  The  line  joining  the  points  of  contact  of  two  parallel 
tangents  is  a  diameter. 

208.  If  two  tangents  have  their  points  of  contact  the  oppo- 
site extremities  of  a  diameter,  the  tangents  are  parallel. 

209.  If  a  radius  bisect  a  chord,  it  will  bisect  the  arc  and  be 
perpendicular  to  the  chord. 

210.  If  a  perpendicular  bisect  a  chord,  it  will,  if  extended, 
pass  through  the  centre  of  the  circle. 

211.  If  a  radius  bisect  an  arc,  it  will  also  bisect  its  sub- 
tending chord  and  be  perpendicular  to  it. 

211  (a) .  If  a  line  bisect  an  arc  and  its  subtending  chord,  this 
line  will,  if  extended,  pass  through  the  centre  of  the  circle. 

212.  If  two  non-intersecting  chords  intercept  equal  arcs, 
they  are  parallel. 

213.  Converse  of  204. 

214.  If,  in  the  same  or  equal  circles,  two  chords  be  equal, 
they  are  at  equal  distances  from  the  centres. 

Sug.   Consult  72,  Remark  (a). 

215.  Converse  of  214. 

216.  If,  in  the  same  or  equal  circles,  two  chords  be  unequal, 
the  less  chord  is  the  farther  from  the  centre. 

Post.  Let  ABD  and  HPK  be  two  equal  circles,  and  chord 
AB  >  chord  HK 

We  are  to  prove  that  HK  is  farther  from  the  centre  N  than 
AB  is  from  the  centre  Q. 


44  PLANE  GEOMETRY. 

Cons.  Draw  the  radius  C  V  perpendicular  to  AB  and  radius 
NZ  perpendicular  to  HK.  Draw  also  the  radii  CA,  CB,  NH, 
and  NK. 

Then  CO  is  the  distance  AB  is  from  C,  and  NS  is  the  dis- 
tance HKis  from  N.  (72,  Eemark  (a).) 

So  we  are,  in  reality,  to  prove  NS  >  CO. 


Dem.  AB  >  HK  Why  ? 

05  is  what  part  of  AB  ?     Why  ? 

ITS  is  what  part  of  HK?     Why? 

Then  what  is  the  relation  of  OB  and  HS  in  respect  to 
magnitude  ? 

Express  that  relation  by  symbols. 

What  is  the  relation  between  the  two  arcs  A  VB  and  HZKt 
Why? 

What  relation,  then,  between  the  A  ACB  and  HNK?  Why  ? 

How,  then,  does  the  sum  of  the  angles  A  and  B  compare 
with  the  sum  of  the  angles  IT  and  N?  Explain. 

How  does  Z  A  compare  with  Z  B ?    ZHwithZK?    Why? 

What  relation,  then,  in  respect  to  magnitude,  between  the 
AH  and  B?  Why? 

Now,  on  BO  mark  off  a  distance  from  B  equal  to  HS,  as 
BT,  and  join  CT. 

Compare  CT  and  NS.     (See  Theorem  123  (a).) 

Now  compare  them  both  with  CO. 

Hence,  etc. 


PLANE  GEOMETRY. 


45 


The  following  variation  of  the  above  method  was  given  by 
one  of  the  author's  pupils. 

Make  AE  =  CD.  Then,  since  HR  is  less  than  PD,  ZAia 
less  than  Z  C.  (Theorem  124.) 


The  rest  is  similar  to  the  one  above. 

217.  Converse  of  216. 

Sug.   Three  possible  relations. 

218.  Through  any  three  points  selected  at  random,  provided, 
however,  they  are  not  in  the  same  straight  line,  one  circum- 
ference can  be  made  to  pass,  and  only  one;  or,  if  more  be 
drawn,  they  must  coincide. 

Sug.  If  the  points  be  joined,  these  lines  must  be  chords 
of  the  circle.  Then  consult  210. 

219.  If  two  unequal  circles  are  concentric,  the  chords  of  the 
greater  which  are  tangents  of  the  less  are  equal. 

Sug.   Consult  201,  215,  and  72,  Remark  (a). 

220.  If  two  circles  be  tangent  to  each  other  externally,  the 
radii  drawn  to  the  point  of  contact  form  one  and  the  same 
straight  line. 

Post.  Let  the  two  circles  PCN  and  CDK  be  tangent  to 
each  other  externally  at  C,  and  AC  and  BC  the  radii  drawn 
to  the  point  of  contact  C. 

We  are  to  prove  that  ACB  is  a  straight  line.  If  we  can 
prove  that  ACB  is  the  shortest  distance  from  A  to  B,  then 
ACB  must  be  a  straight  line.  For  if  a  straight  line  be  drawn 


46  PLANE  GEOMETRY. 

from  A  to  B,  it  will  be  the  shortest  distance  between  A  and  B. 
(Sect.  42,  XIV.)  And  since  there  can  be  but  one  shortest 
distance,  if  ACB  is  proved  also  to  be  the  shortest  distance, 
then  ACB  must  coincide  with  that  line,  and  consequently  be 
a  straight  line. 

Jf 


So  we  are  to  prove  that  ACB  is  the  shortest  distance  from 
-4  to  B. 

Cons.  Draw  any  other  radius  than  BC  in  circle  CDK,  as 
BD,  and  join  AD. 

Dem.  Since  every  point  in  the  circumference  CDK  except 
C  is  outside  the  circumference  CPN  (see  Def.),  AD  must 
extend  beyond  the  circumference,  and 

AD  >  AC.  Why  ? 

DB=CB.  Why? 


DB>AC+CB.  Why? 

But  D  is  any  point  in  the  circumference  CDK  except  C. 
Hence,  the  distance  from  A  to  B  by  way  of  D  is  always  greater 
than  by  way  of  C.  Or  ACB  is  the  shortest  distance  between 
A  and  B,  and  is  therefore  a  straight  line.  Q.E.D. 

221.  If  radii  be  drawn  to  the  point  of  contact  of  two  circles 
that  are  tangent  to  each  other  externally,  and  a  straight  line 
be  drawn  through  the  point  of  contact  perpendicular  to  one  of 
the  radii,  this  line  will  be  a  common  tangent  to  the  two  circles. 


PLANE  GEOMETRY.  47 

222.  If  two  circles  be  tangent  to  each  other  externally,  the 
straight  line  joining  their  centres  will  pass  through  the  point 
of  contact. 

223.  If  two  circles  be  tangent  to  each  other  internally,  and  a 
line  be  drawn  through  the  point  of  contact  tangent  to  the 
outer  circle,  it  will  also  be  tangent  to  the  inner  circle. 

224.  If  two  circles  be  tangent  to  each  other  internally,  and 
radii  be  drawn  to  point  of  contact,  these  radii  will  lie  in  the 
same  straight  line. 

225.  If  two  circles  be  tangent  to  each  other  internally,  the 
line  joining  their  centres  will,  if  extended,  pass  through  the 
point  of  contact. 

226.  If  two  circles  be  tangent  to  each  other  internally,  and 
a  common  tangent  be  drawn  through  the  point  of  contact,  a 
perpendicular  to  this  tangent  at  the  point  of  contact  will,  if 
extended,  pass  through  the  centres  of  both  circles. 

227.  If  two  circles  be  tangent  to  each  other  internally,  the 
radius  of  the  smaller  to  the  point  of  contact  will,  if  extended, 
pass  through  the  centre  of  the  larger,  and  the  radius  of  the 
larger  to  point  of  contact  will  also  pass  through  the  centre  of 
the  smaller. 

228.  If  two  circles  be  tangent  to  each  other  internally,  and 
a  line  be  drawn  from  the  centre  of  either  circle  perpendicular 
to  their  common  tangent,  this  perpendicular  will  pass  through 
the  centre  of  the  other  circle  and  also  through  the  point  of 
contact. 

229.  Two  unequal  circles  may  have  five  positions  relative  to 
each  other,  viz. : 

I.  One  may  lie  wholly  within  the  other  without  contact. 

II.  They  may  be  tangent  to  each  other  internally. 

III.  They  may  intersect  each  other. 

IV.  They  may  be  tangent  to  each  other  externally. 

V.  One  may  lie  wholly  without  the  other  without  contact. 


48  PLANE  GEOMETRY. 

If  the  circles  are  equal,  how  many  positions  relative  to  each 
other  may  they  have  ? 

In  Case  I.  what  relation  exists  between  the  line  joining 
their  centres  and  the  sum  of  their  radii  ?  Between  the  same 
line  and  the  difference  of  their  radii  ?  Prove. 

Prove  both  the  above  relations  in  each  of  the  other  four 
cases. 

What  must  be  the  position  of  two  circles,  then,  if  the  dis- 
tance between  their  centres  is 

Jf.  ft? 

II.  Less  than  the  sum,  and  also  less  than  the  difference,  of 
their  radii  ? 

III.  Equal  to  the  difference  of  their  radii  ? 

IV.  Less  than  the  sum,  and  greater  than  the  difference,  of 
their  radii  ? 

V.  Equal  to  the  sum  of  their  radii  ? 

VI.  Greater  than  the  sum  of  their  radii  ? 

230.  If  two  circles  intersect  each  other,  the  line  joining 
their  centres  will  bisect  their  common  chord  and  be  perpen- 
dicular to  it. 

231.  If  two  circles  intersect  each  other,  and  radii  be  drawn 
to  the  middle  point  of  their  common  chord,  these  radii  will 
form  one  and  the  same  straight  line. 

232.  If  two  circles  intersect  each  other,  and  either  radius 
be  drawn  bisecting  their  common  chord,  this  radius  will,  if 
extended,  pass  through  the  centre  of  the  other  circle. 

233.  If  two  circles  intersect  each  other,  and  a  perpendicular 
bisect  their  common  chord,  this  perpendicular  will,  if  extended, 
pass  through  the  centres  of  both  circles. 

234.*  A  central  angle  is  measured  by  the  arc  which  its  sides 
intercept  on  the  circumference. 

*  See  Appendix. 


PLANE  GEOMETRY. 


49 


Dem.  In  the  circle  ANQ  let  us  conceive  the  radius  CA 
to  move  about  C  as  a  pivot,  remaining  always  in  the  same 
plane.  In  one  complete  revolution  it  is  evident  that  the 
extremity  A  will  have  traced  the  entire  circumference.  Like- 
wise, in  one-half  a  revolution, 
as  when  it  has  reached  the 
position  CP}  making  PGA  a 
diameter,  the  extremity  A  will 
have  traced  one-half  the  circum- 
ference. So  when  it  has  reached 
the  position  CN  perpendicular 
to  AP,  it  will  have  traced  one- 
fourth  the  circumference  (see 
199,  and  59  and  60).  In  like 
manner,  when  it  has  reached 
the  position  CK,  making  the 
angles  ACK  and  KCN  equal, 

then  tne  arcs  AK  and  KN  are  also  equal  (see  191) ;  that  is, 
the  arc  AK  is  one-eighth  of  the  circumference,  and  the  angle 
ACK  is  one-eighth  of  the  angular  space  about  the  point  C. 
In  like  manner,  the  arc  AH  is  one-sixteenth,  AD  one  thirty- 
second,  AB  one  sixty-fourth,  etc.,  of  the  entire  circumference, 
while  their  corresponding  central  angles  are  the  same  parts  of 
four  right  angles,  or  the  angular  space  about  the  centre  (7. 
So  that,  whatever  the  position  of  the  radius  CA,  the  arc 
described  will  be  the  same  part  of  the  entire  circumference 
that  its  corresponding  central  angle  is  of  the  angular  space 
about  the  centre.  If,  now,  the  entire  circumference  be  divided 
into  360  equal  parts,  each  part  would  correspond  to  a  central 
angle  of  1°.  Consequently,  if  an  arc  contain  27  of  the  360 
equal  parts,  it  may  be  called  an  arc  of  27°,  and  its  correspond- 
ing central  angle  would  contain  -ff^  of  four  right  angles,  or 
would  be  an  angle  of  27°.  Likewise,  if  the  arc  contains  a 
whole  number  of  degrees  and  a  fraction,  the  latter  can  be 
expressed  in  the  smaller  units.  For  example,  suppose  the  arc 


50 


PLANE  GEOMETRY. 


to  contain  48°  37'  24.92" ;  then  the  corresponding  central  angle 
would  contain  the  same  number  of  angular  units ;  i.e.  it  would 
be  an  angle  of  48°  37'  24.92",  to  the  same  degree  of  approxima- 
tion. Therefore  the  arc  is  said  to  measure  the  corresponding 
central  angle,  because,  as  above  stated,  it  is  the  same  part  of 
the  entire  circumference  that  its  corresponding  central  angle  is 
of  the  angular  space  about  the  centre.  Q.E.D. 

235.  An  inscribed  angle  is  measured  by  one-half  the  arc 
intercepted  by  its  sides. 

Remark.  In  demonstrating  this  theorem  it  will  simplify  it 
somewhat  to  make  three  cases  of  it ;  viz. : 

I.  When  one  of  the  sides  of  the  angle  is  a  diameter. 

II.  When  the  centre  is  between  the  sides  of  the  angle. 

III.  When  the  centre  is  without  the  angle. 

K 


D 


Fig.  I. 

CASE  I.  —  Post.  Let  B  be  an  inscribed  angle  with  one  of 
its  sides,  as  AB,  a  diameter. 

We  are  to  prove  that  the  angle  B  is  measured  by  one-half 
the  arc  AD;  that  is,  one-half  the  arc  AD  is  the  same  part 
of  the  entire  circumference  ABD  that  the  angle  B  is  of  the 
angular  space  about  B. 

Cons.   Draw  the  radius  CD  (Fig.  I.). 

Dem.   What  relation  exists  between  CD  and  CB  ?    Why  ? 

What  relation,  then,  between  angles  B  and  D  ?    Why  ? 


PLANE  GEOMETRY.  51 

What  relation  exists  between  the  sum  of  the  angles  B  and 
D  and  angle  DCA?  Why? 

What  relation,  then,  exists  between  the  angle  B  and  the 
angle  DCA?  Whj? 

What  measures  the  angle  DCA  ?     Why  ?     (Consult  234.) 

What,  then,  must  measure  the  angle  B  ? 

Again,  Fig.  II.,  Case  I. 

Cons.   Divaw  a  diameter,  as  UK,  parallel  to  BD. 

Dem.  What  relation  exists  between  the  angles  BCK  and 
HCA  ?  Why  ? 

What  relation,  then,  .exists  between  the  two  arcs  HA  and 
BK?  Why?  (See  191.) 

What  relation  between  the  two  arcs  DH and  BK?     Why? 

What  relation,  then,  between  the  two  arcs  DH  and  AH? 
Why? 

What  relation,  then,  between  the  two  arcs  AH  and  AD  ? 
Why? 

What  relation  between  the  two  angles  HCA  and  B  ?    Why  ? 

What  measures  the  angle  HCA  ?     Why  ? 

What,  then,  must  measure  the  angle  B  ? 

After  answering  correctly  the  foregoing  questions  the  pupil 
should  have  no  difficulty  in  writing  "out  a  complete  demonstra- 
tion of  each  of  the  three  cases,  using  Case  I.  in  demonstrating 
II.  and  III. 

236.  If  two  angles  be  inscribed  in  the  same  or  equal  circles, 
and  their  sides  intercept  the  same  or  equal  arcs,  the  two  angles 
are  equal. 

236  (a).   Converse  of  236. 

236  (6).  What  part  of  the  circumference  measures  a  right 
angle  ?  Prove. 

What  relation  exists  between  angles  inscribed  in  the  same 
segment  ?  Why  ? 

What  measures  an  angle  inscribed  in  a  semicircle  ? 

What  kind  of  an  angle,  then,  must  it  be  ? 


52  PLANE  GEOMETRY. 

If  an  angle  be  inscribed  in  a  segment  less  than  a  semicircle, 
what  kind  of  an  angle  must  it  be  ?  Why  ? 

If  an  angle  be  inscribed  in  a  segment  greater  than  a  semi- 
circle, what  kind  of  an  angle  must  it  be  ?  Why  ? 

237.  If  an  angle  be  formed  by  two  chords  whose  vertex  is 
between  the  centre  and  circumference,  it  will  be  measured  by 
one-half  the  sum  of  its  two  intercepted  arcs. 

238.  If  the  vertex  of  an  angle  formed  by  two  secants  is 
without  the  circle,    this  angle  is  measured  by  one-half  the 
difference  of  the  two  intercepted  arcs. 

239.  If  an  angle  be  formed  by  a  tangent  and  chord,  this 
angle  is  measured  by  one-half  the  arc  intercepted  by  its  sides. 

240.  If  the  vertex  of  an  angle  formed  by  a  tangent  and 
secant  is  without  the  circle,  this  angle  is  measured  by  one-half 
the  difference  of  the  two  intercepted  arcs. 

241.  If  an  angle  be  formed  by  two  tangents  to  the  same 
circle,  this  angle  is  measured  by  one-half  the  difference  of  the 
two  intercepted  arcs. 

242.  If  from  the  same-  point  without  a  circle  two  tangents 
be  drawn,  these  two  tangents  are  equal;   i.e.  the  distances 
from  the  common  point  to  the  points  of  contact  are  equal. 

Sug.   Consult  either  201  and  107,  or  239  and  109. 

ADVANCE  THEOREMS. 

243.  The  line  which  joins  the  vertex  of  an  angle  formed 
by  two  tangents  to  the  centre  of  the  circle  bisects  the  angle, 
and  also  the  chord  which  joins  the  points  of  contact. 

244.  If,  from  the  same  point,  two  tangents  to  a  circle  be 
drawn  whose  points  of  contact  are  the  extremities  of  a  chord, 
the  angle  formed  by  the  two  tangents  is  double  the  angle 
formed  by  the  chord  and  diameter  drawn  from   either   ex- 
tremity of  the  chord, 


PLANE  GEOMETRY.  53 

245.  If  two  circles  which  are  tangent  to  each  other  externally 
have  three  common  tangents,  the  one  that  passes  through  the 
point  of  contact  of  the  two  circles  bisects  the  other  two. 

Sug.    Consult  242. 

246.  If  four  tangents  form  a  quadrilateral,  the  sum  of  two 
opposite  sides  equals  the  sum  of  the  other  two  opposite  sides. 

247.  If  two  mutually  equiangular  triangles  be  inscribed  in 
equal  circles,  the  triangles  are  equal  in  all  respects. 

248.  If  two  opposite  sides  of  an  inscribed  quadrilateral  are 
equal,  the  other  two  sides  are  parallel. 

248  (a).   Converse  of  248. 

249.  The  opposite  angles  of  an  inscribed  quadrilateral  are 
supplementary. 

250.  If  two  equal  chords  be  drawn  from  opposite  extremities 
of  a  diameter  and  011  opposite  sides  of  it,  they  will  be  parallel. 

251.  If  two  chords  be  drawn  through  the  same  point  in  a 
diameter  making  equal  angles  with  it,  they  are  equal. 

252.  If  one  of  the  equal  sides  of  an  isosceles  triangle  be  the 
diameter  of  a  circle,  the  circumference  will  bisect  the  base. 

253.  If  an  equilateral  triangle  be  inscribed  in  a  circle,  and 
a  diameter  be  drawn  from  one  vertex,  the  triangle,  formed  by 
joining  the  other  extremity  of  the  diameter  and  the  centre  of 
the  circle  with   one  of  the  other  vertices   of  the  inscribed 
triangle,  will  also  be  equilateral. 

254.  The  bisectors  of  the  angles  formed  by  extending  the 
sides  of  an  inscribed  quadrilateral  are  perpendicular  to  each 
other. 

255.  If  an  equilateral  triangle  be  inscribed  in  a  circle,  and 
any  point  in  the  circumference  be  selected  at  random,  one  of 
the  lines  which  join  this  point  to  the  three  vertices  will  equal 
the  sum  of  the  other  two. 


54  PLANE  GEOMETRY. 

256.  If  an  inscribed  isosceles  triangle  have  its  base  angles 
each  double  the  vertical  angle,  and  its  vertices  be  the  points 
of  contact  of   three    tangents,  these  tangents  will   form   an 
isosceles  triangle  each  of  whose  base  angles  is  one-third  its 
vertical  angle. 

257.  If  through  any  point  selected  at  random  in  a  circle  a 
diameter  and  other  chords  be  drawn,  the  least  chord  will  be 
the  one  that  is  perpendicular  to  the  diameter. 

258.  ADB  is  a  semicircle  of  which  the  centre  is  (7,  and 
AEG  is  another  semicircle  on  the  diameter  AC,  and  AT  is  a 
common  tangent  to  the  two  semicircles  at  A.     Prove  that,  if 
from  any  point  F  in  the  circumference  of  the  first  a  straight 
line  be  drawn  to  (7,  the  part  FK,  cut  off  by  the  second  semi- 
circle, is  equal  to  FH,  perpendicular  to  the  tangent  AT. 

259.  If  a  triangle  ABO  be  formed  by  the  intersection  of 
three  tangents  to  a  circumference  whose  centre  is  0,  two  of 
which,  AM  and  AN,  are  fixed,  while  the  third,  BGj  touches 
the  circumference  at  a  variable  point  P,  prove 

I.  that  the  perimeter  of  the  triangle  is  constant. 

II.  that  the  angle  BOO  is  constant. 

260.  If  the  sides  of  any  quadrilateral  be  the  diameters  of 
circles,  the  common  chord  of  any  adjacent  two  is  parallel  to 
the  common  chord  of  the  other  two. 

EATIO  AND  PBOPOKTION, 

(It  is  assumed  that  the  pupil  is  familiar  with  the  algebraic 
processes  involved  in  the  manipulation  of  equations  of  the 
first  and  second  degree.  For  definition  of  Geometrical  Magni- 
tudes, see  37.) 

261.  If  of  two  unequal  magnitudes  the  less  is  contained  an 
exact  number  of  times  in  the  greater,  the  latter  is  said  to  be  a 
multiple  of  the  former,  and  the  former  an  aliquot  part  of  the 
latter. 


PLANE  GEOMETRY.  55 

262.  If  of  three  unequal  magnitudes  the  smallest  is  con- 
tained an  exact  number  of  times  in  each  of  the  two  larger,  the 
latter  are  said  to  be  commensurable,  and  the  former  is  said  to 
be  their  common  measure. 

263.  When  no  magnitude  can  be  found  which  is  contained 
an  exact  number  of  times  in  each  of  two  magnitudes,  the  latter 
are  said  to  be  incommensurable. 

264.  If  two  commensurable  magnitudes  contain  their  com- 
mon measure,  one  n  times  and  the  other  m  times,  they  are  said 
to  be  to  each  other  as  n  to  m,  or  in  the  ratio  of  n  to  m. 

x 
A 
B 


For  example,  if  the  line  x  is  contained  in  the  line  A  3  times, 
and  in  line  B  5  times,  then  A  is  to  B  as  3  to  5,  and  the  line  x 
is  a  common  measure  of  the  two  lines  A  and  B. 

265.  Equimultiples  of  two  or  more  magnitudes  are  the  results 
obtained  by  multiplying  these  magnitudes  by  the  same  number. 
(See  39.) 


Thus,  if  x  and  y  be  any  two  lines,  and  a  and  b  two  other 
lines  such  that  the  former  contains  the  line  x  n  times  (in  this 
particular  case  n  is  3),  and  the  latter  contains  the  line  y 
n  times,  then  a  and  b  are  equimultiples  of  x  and  y  respectively. 

266.  If  two  magnitudes  are  to  each  other  as  m  to  n  (see  264), 
it  is  usually  expressed   thus,    m:n,    called  the  ratio   form; 

or  thus,  ~,  called  the  fractional  form, 
n 

267.  Hence  a  ratio  may  be  defined  as  an  expression  of  com- 
parison, in  respect  to  size,  of  two  magnitudes  of  the  same 
kind. 


56  PLANE  GEOMETRY. 

268.   If  x  and  y  are  two  magnitudes,  and  y  a  multiple  of  x 


(see  261),  the  latter  being  contained  in  the  former  n  times, 
then  the  ratio  of  y  to  x  is  as  n  to  unity,  or  n  :  1. 

269.  If  x  and  t/  are  commensurable  (262),  and  their  common 
measure  is  contained  n  times  in  a?,  and  m  times  in  y,  then  the 
ratio  of  x  to  y  is  as  n  to  ra,  or,  technically  expressed,  n :  m. 

270.  If  x  and  ?/  are  incommensurable  (see  263),  the  ratio 
cannot  be  exactly  expressed.     If,  however,  x  and  y  are  num- 
bers, the  ratio  may  be  approximately  expressed  by  placing  1 
for  the  first  term,  and  the  quotient  of  the  greater  divided  by 
the  less  to  any  number  of  decimal  places  for  the  second  term ; 
thus,  l:n.pgr£-K     In  such  a  case  the  computation  can  be 
carried  to  any  required  degree  of  approximation. 

271.  In  a  similar  way  can  be  obtained  an  approximate  ratio 
of  two  incommensurable  magnitudes,  expressed  numerically; 


If  x  and  y  be  two  incommensurable  magnitudes,  suppose  x 
to  be  divided  into  a  certain  number,  M,  of  equal  parts,  and  that, 
on  applying  one  of  these  equal  parts  to  y,  it  is  found  to  be 
contained  m  times  with  a  remainder  less  than  this  part.  Then 
it  is  evident  that  the  ratio  of  y  to  x  is  neither  m :  w,,  nor 
m  +  l:n,  but  is  greater  than  the  former  and  less  than  the 
latter;  i.e.  greater  than  the  value  of  — ,  and  less  than  the 

value  of  m  "*"   ,  or  —  -| —     It  is  thus  seen  that  the  smaller 
n  n      n 

the  unit  into  which  x  is  divided,  the  greater  becomes  the  value 
of  n,  and  consequently  the  smaller  becomes  the  value  of  -. 
And  hence,  neglecting  this  value,  —  will  express  the  approxi- 


PLANE  GEOMETRY.  57 

mate  ratio  of  y  to  x,  the  degree  of  approximation  depending 
upon  the  size  of  the  unit  of  measure. 

272.  The  first  term  of  a  ratio  is  called  the  antecedent,  and 
the  second  term  the  consequent. 

273.  A  proportion  is  an  expression  of  equality  between  two 
equal  ratios,  usually  indicated  by  four  dots  between  the  two 
ratios  ;  thus,  a  :  b  :  :  x  :  y,  and  read,  a  is  to  b  as  x  is  to  y. 

274.  The  four  magnitudes  forming  a  proportion  are  called 
proportionals. 

275.  If  a  proportion  contains  only  two  ratios,  it  is  called  a 
simple  proportion ;  if  more  than  two  and  all  equal,  it  is  called 
a  continued  proportion. 

276.  The  first  and  last  terms  of  a  simple  proportion  are 
called  the  extremes,  and  the  other  two  the  means. 

277.  In  a  simple  proportion,  where  the  terms  are  all  of 
different  values,  each  term  is  said  to  be  a  fourth  proportional 
to  the  other  three. 

278.  In  a  simple  proportion,  where  the  two  means  or  the  two 
extremes  are  alike,  the  repeated  quantity  is  said  to  be  a  mean 
proportional  to  the  other  two,  and  the  proportion  is  called  a 
mean  proportion. 

279.  In   a  mean  proportion,  either  of  the  quantities  not 
repeated  is  said  to  be  a  third  proportional  to  the  other  two. 

280.  When  a  line  is  so  divided  that  the  larger  part  is  a 
mean  proportional  between  the  whole  line  and  the  smaller 
part,  it  is  said  to  be  divided  in  extreme  and  mean  ratio. 

281.  It  will  be  found  on  investigation  that  the  order  of  four 
proportionals,  when  all  of  the  same  kind,  can  be  varied,  as  well 
as  certain  other  transformations  effected,  without  destroying 
the  proportion.     The  principal  of  these  are  as  follows  : 


58  PLANE  GEOMETRY. 

282.  Magnitudes  are  said  to  be  in  proportion  by  alternation 
when  either  the  two  means  or  the  two  extremes  are  made  to 
exchange  places. 

283.  Magnitudes  are  said  to  be  in  proportion  by  inversion 
when  the  means  are  made  to  exchange  places  with  the  extremes. 

284.  Magnitudes  are  said  to  be  in  proportion  by  composition 
when  the  sum  of  the  terms  of  the  first  ratio  is  to  either  term 
of  the  first  ratio  as  the  sum  of  the  terms  of  the  second  ratio 
is  to  the  corresponding  term  of  that  ratio. 

285.  Magnitudes  are  said  to  be  in  proportion  by  division 
when  the  difference  of  the  terms  of  the  first  ratio  is  to  either 
term  of  that  ratio  as  the  like  difference  of  the  terms  of  the 
second  ratio  is  to  the  corresponding  term  of  that  ratio. 

286.  Magnitudes  are  said  to  be  in  proportion  by  composition 
and  division  when  the  sum  of  the  terms  of  the  first  ratio  is  to 
their  difference  as  the  sum  of  the  terms  of  the  last  ratio  is  to 
their  like  difference. 

THEOREMS. 

287.  Equimultiples  of  two  magnitudes  are  in  the  same  ratio 
as  the  magnitudes  themselves. 

CASE  I.  —  When  the  magnitudes  are  commensurable. 

a    \ i i .' 

b    I i 1 5 


Let  x  and  y  be  two  commensurable  magnitudes,  and  a  and  b 
be  their  respective  equimultiples. 

We  are  to  prove  x  :  y  :  a  :  b. 

Dem.  Since  x  and  y  are  commensurable  (262),  they  must 
have  a  common  measure,  as  the  line  c.  Suppose  it  to  be  con- 
tained in  x  n  times,  and  in  y  m  times ;  i.e.  that  on  dividing 


PLANE  GEOMETRY.  59 

the  two  lines  x  and  y  into  n  and  m  equal  parts  respectively, 
the  parts  will  all  individually  be  equal  to  the  line  c.  Con- 
sequently if  the  line  c  be  multiplied  first  by  n  and  then  by  m 
(see  39),  the  resulting  lines  will  be  exactly  equal  in  length  to 
the  lines  x  and  y  respectively. 

These  relations  may  be  symbolically  expressed  as  follows: 

(1)  -  =  c    and          ^  =  c. 
n  m 

(2)  n  x  c  =  x    and    m  x  c  =  y. 

n  X  c      x  n      x 

Whence  —  =  -,    or  —  =  -. 

m  x  c     y  my 

Let  q  be  the  number  by  which  x  and  y  are  multiplied  to 
obtain  the  equimultiples  a  and  b ;  then 

(3)  q  x  x  =  a      and      q  xy  =  b. 

But  from  (2)  g  x  x  =  q  x  ft  X  c, 

and  qxy  =  qxmxc. 

Whence  a  =  q  xnx  c, 

and  b  —  qxmxc. 

a      a  x  n  x  c      n 

Therefore  -  =  i— =  — 

b      q xmx c     m 

Whence  -  =  -,   or   a:b::x:y.  Q.E.D. 

b     y 

CASE  II.  —  When  the  magnitudes  are  incommensurable. 


Let  x  and  y  be  two  incommensurable  magnitudes,  and  a  and 
b  two  equimultiples  of  x  and  y. 


60  PLANE  GEOMETRY. 

We  are  to  prove  y  :  x  : :  b  :  a. 

Let  x  be  divided  into  n  equal  parts,  and  designate  the  value 
of  one  of  these  parts  by  p.  Let  q  denote  the  number  of  times 
x  and  y  are  contained  in  a  and  b  respectively.  Since  x  and  y 
are  incommensurable,  p  will  be  contained  in  y  m  times  with  a 
remainder  r. 

Then  x  =  np  (1)  and       qx  =  qnp ; 

but  a  =  qx,  .*.    a  =  qnp.  (2) 

Again,  y  =  mp  +  r  (3)  and       qy  =  qmp  -f  qr ; 

but  b—qy,  .'.    b  =  qmp  +  qr.  (4) 

Dividing  (4)  by  (2),  and  (3)  by  (1), 

•  b  _  qmp       qr  y_mpr 

a~~  qnp      qnp  x  ~  np      np' 

Simplifying, 

&  =  ™  +  Jl  and        ?  =  ™  +  ±. 

a      n      np  x      n      np 

Whence       -  =  -,     or    y  :  x  : :  b  :  a.  Q.E.D. 

x     a 

288.  If  two  commensurable  ratios  can  be  expressed  by  the 
same  numerical  value,  then  these  two  ratios  are  equal,  and 
consequently  the  four  magnitudes  composing  those  ratios  are 
proportionals. 

Let  a,  b,  x,  and  y  represent  four  magnitudes,  a  and  b  being 
commensurable,  also  x  and  y,  so  that,  on  applying  their  re- 

,.  .        „  a     m          x     m 

spective  units  of  measure,  =-  =  —  and  -  =  — 

b      n  y      n 

n       o* 
Then  r  =  -  (Axiom  III.)  or  a:  b::  x:  y.  Q.E.D. 


PLANE  GEOMETRY.  61 

289.  If  two  incommensurable  ratios  can  be  expressed  by  the 
same  approximate  numerical  value,  however  small  the  unit  of 
measure,  then  these  two  ratios  are  equal,  and  the  four  magni- 
tudes comprising  those  ratios  are  proportionals. 

Post.   Let  a :  b  and  x :  y  be  two  ratios,  each  of  which  is 
incommensurable  and  whose  true  values  lie  between  the  ap- 
proximate values  —  and  — u  — .     (See  271.) 
n  n      n 

We  are  to  prove  that  a:b::x:y;  i.e.  that  there  is  no  dif- 
ference between  the  values  of  -  and  - ;  i.e.  that  they  are  equal. 

b          y 

Dem.  It  is  evident  that  -  and  -  are  either  equal  or  unequal. 
b          y 

Let  us   assume  that  they  are  unequal,  and  designate  their 
difference   by  d.      This   difference   being  fixed   and   definite 
cannot  be  changed  by  any  legitimate  manipulation  or  trans- 
formation of  the  ratios. 
By  the  hypothesis, 

(1)  ^>™  and     ?>™. 
b      n  y      n 

(2)  ?<!?  +  l     and     2<5  +  i. 
b     n     n  y      n     n 

Consequently  the  difference  between  -  and  -  must  be  less 
-,  by 

than  ±. 
n 

Now  by  decreasing  the  value  of  the  common  unit  of  measure, 
the  value  of  n  increases  and  that  of  -  decreases  accordingly. 

Hence  we  may  conceive  n  to  be  so  great  that  -  will  be  smaller 

n 

than  d ;  i.e.  the  difference  between  ?  and  ~  is  equal  to  d  and 

b         y 

less  than  d  at  the  same  time.  This  is  of  course  absurd  or 
impossible.  Hence  they  cannot  differ  in  value ;  i.e.  they  are 
equal,  and  a:b::x:y.  Q.E.D. 


62  PLANE  GEOMETRY. 

290.  In  the  demonstration  of  the  following  theorems,  the 
expression  "  four  quantities  "  means  the  numerical  measures 
of  four  geometrical  magnitudes  of  the  same  kind.     It  would 
be  well,  however,  to  represent  the  magnitudes  either  by  angles 
or  lines,  as  in  287. 

291.  If  four  quantities  form  a  proportion,  the  product  of  the 
means  equals  the  product  of  the  extremes. 

292.  Remark.   Our  ability  to  demonstrate  the  theorems  in 
proportion  depends  on  our  knowledge  of  the  principles  govern- 
ing the  manipulation  of  the  equation.    Hence  the  first  thing  to 
be  done  is  to  change  the  proportion  form  to  the  equivalent 
equation  form.    The  author  is  of  the  firm  belief  that  mathe- 
maticians  have  no  right  to  amalgamate   these  two  forms  of 
expression,  and  that  pupils  should  be  taught  to  rigidly  dis- 
criminate in  their  use. 

Post.   Let  the  four  quantities  a,  6,  c,  and  d  form  a  propor- 
tion so  that 

a  :  b  :  :  c  :  d. 

We  are  to  prove  that       ad  =  be. 

Dem.  a-  =  * 

b     d 

(Changing  to  equation  form.) 


(Multiplying  both  members  of  the  equation  by  d.) 
ad  =  be. 

(Multiplying  both  members  of  the  previous  equation  by  b.) 

Q.E.D. 

293.   If  three  quantities  form  a  proportion,  the  product  of 
the  extremes  is  equal  to  the  square  of  the  mean. 


PLANE  GEOMETRY.  63 

294.  If  the  product  of  two  quantities  equals  the  product 
of  two  others,  the  two  factors  of  either  product  may  be  made 
the  means,  and  the  other  two  the  extremes  of  a  proportion. 

Post.   Let  ex  =  an. 

We  are  required  to  form  a  proportion  from  the  four  factors 
c,  x,  a,  and  n,  in  which  c  and  x  shall  be  the  extremes. 

Dem.  c  =  ™.  Why? 

°  =  1  Why? 

a     x 

.*.     c  :  a  : :  n  :  x. 
(Changing  from  equation  form  to  the  proportion  form.) 

Form  proportions  from  the  following  equations  in  which  the 
factors  of  the  product  marked  Ex.  shall  form  the  extremes, 
avoiding  the  factor  1. 


Ex. 

I. 

2  x  =  ac. 

Ex. 

II. 

6  =  ax. 

Ex. 

III. 

3  Va  =  14. 

Ex. 

IV. 

a(x  +  y)  =  n2. 

Ex. 

V. 

2/2=  an  -f-  ac. 

Ex. 

VI. 

ab  +  bx  =  en  + 

yc. 

Ex. 

VII. 

aic  +  aj  =  2/62  + 

b2. 

Ex. 

VIII. 

a2—  1  =  a2—  62. 

Ex. 

IX. 

ax  +  xb  +  cx  = 

nd  +  7m  +  wfc. 

Ex. 

X. 

x*+2cx  +  c?= 

an  +  ny. 

64  PLANE  GEOMETRY. 

295.   If  four  quantities  form  a  proportion,  they  will  be  in 
proportion  by  inversion. 

.   Divide  1  by  each  member  of  the  equation. 


296.   If  four  quantities  form  a  proportion,  they  will  be  in 
proportion  by  alternation. 

Post.   Let  the  four  quantities  a,  a,  c,  and  n  form  a  propor- 
tion so  that 

a  :  x  :  :  c  :  n. 
We  are  to  prove  that 

a  :  c  :  :  x  :  n. 

Dem.  -  =  -.  Why  ? 

x     n 

-  =  -.  Why? 

xc     n 

Why? 


.*.     a  :  c  : :  x  :  n.  Q.E.D. 

Query.   If  the  corresponding  terms  of  two  proportions  be 
added,  will  the  sums  form  a  proportion  ?     Numerical  example. 

297.   If  four  quantities  form  a  proportion,  they  will  be  in 
proportion  by  composition. 

Post.   Let  the  four  quantities  x,  y,  a,  and  b  form  a  pro- 
portion so  that 

x  :  y  :  :  a  :  b. 

We  are  to  prove, 

I.  x  -\-y  :y  : :  a  -\-b  :b. 

II.  x  -\-  y  :  x  : :  a  +  b  :  a. 

Dem.  -  =  -.  Why? 

y     b 


PLANE  GEOMETRY.  65 

+1  =     +  1.  Why? 


~  +  y-  =  l  +  l  Why? 

y     y      b^  b 


-.  Why? 

y         b 

.:     x  +  y:y::a  +  b:b.  Why? 

Q.E.D. 

The  pupil  should  demonstrate  Part  II. 
Sug.   Consult  295. 

298.  If  four  quantities  form  a  proportion,  they  will  be  in 
proportion  by  division. 

Sug.  There  are  four  cases.  In  Case  I.  subtract  1  from  each 
member  of  the  equation.  The  other  cases  may  be  similarly 
demonstrated  by  reversing  the  above  process,  and  consulting 
295. 

299.  If  two  proportions  have  a  ratio  in  each  equal,  the  other 
two  ratios  will  form  a  proportion. 

300.  If  two  proportions  have  the  two  antecedents  of  one 
equal  respectively  to  the  two  antecedents  of  the  other,  the 
consequents  will  form  a  proportion. 

301.  If  two  proportions  have  the  two  consequents  of  one 
equal  respectively  to  the  two  consequents  of  the  other,  the 
antecedents  will  form  a  proportion. 

302.  If  four  quantities  form  a  proportion,  they  will  be  in 
proportion  by  composition  and  division. 

Sug.   Consult  298,  297,  296,  and  299. 

303.  If  any  number  of  magnitudes  of  the  same  kind  form  a 
proportion,  the  sum  of  the  antecedents  is  to  the  sum  of  the 
consequents  as  any  antecedent  is  to  its  consequent. 


66  PLANE  GEOMETRY. 

Post.   Let  the  quantities  a,  x,  c,  w,  d,  and  r  form  a  continued 
proportion,  so  that 

a  :  x: :  c:n:  :d:r. 
We  are  to  prove, 


or  II.  :  :  c  :  n  ; 

or  III.  :  :  d  :  r. 

Dem.  Now,  if  this  theorem  can  be  demonstrated,  then 
Case  I.  must  be  a  true  proportion.  Let  us  temporarily  assume 
it  to  be  true,  and  trace  the  results. 

First.  x(a  +  c  +  d)  =  a(x  +  n+r),  Why? 

or  ax  -\-  ex  +  dx  =  ax  -f  an  +  ar  ;  Why  ? 

but  (1)     ax  =  ax.  Why? 

.•.     ex  +  dx  =  an  +  ar.  Why  ? 

(2)      ca;  =  an.  Why  ? 

.-.   (3)      dx  =  ar.  Why? 

Now  it  is  evident  that  if  we  could  obtain  the  equations  (1), 
(2),  (3)  from  our  given  proportions,  we  could  reverse  the 
above  process,  and  thus  demonstrate  the  theorem. 

Let  us  bear  in  mind  that  the  ratio  a  :  x  in  Case  I.  is  the 
one  that  is  to  form  the  proportion  with  that  composed  of  the 
sums  of  the  antecedents  and  consequents,  and  that  equation  (1) 
is  formed  from  the  product  of  the  terms  of  that  ratio. 

Hence                      (1)     ax  =  ax  ;  Why  ? 

(2)  ex  =  an  ;  Why  ? 

(3)  dx  =  ar.          •  Why? 
Hence              ax  -f-  ex  +  dx  =  ax  -f  an  -f-  ar,  Why  ? 

or  x(a  +  c  +  d)  =  a(x  +  n  +  r).  Why? 


The  pupil  should  be  able  to  finish  this  case,  and  also  demon- 
strate the  other  two  without  difficulty.     Consult  204. 


PLANE  GEOMETRY.  67 

304.  If  four  quantities   form  a  proportion,   the  terms  of 
either  ratio  may  be  either  multiplied  or  divided  by  the  same 
quantity,  and  the  results  still  form  a  proportion. 

304  (a).  If  the  antecedents  or  consequents  of  a  proportion 
be  either  multiplied  or  divided  by  the  same  quantity,  the 
results  will  still  form  a  proportion. 

305.  If  four  quantities  form  a  proportion,  and  the  terms  of 
one  ratio  be  either  multiplied  or  divided  by  the  same  quantity, 
while  both  terms  of  the  other  ratio  be  either  multiplied  or 
divided,  either  by  the  same  or  different  quantity  from  that 
used  in  the  first  ratio,  the  results  will  still  form  a  proportion. 

Sug.  The  pupil  should  take  each  case  involved  in  the  state- 
ment of  the  above  theorem  separately. 

306.  If  two  proportions  be  given,  the  products  of  the  corre- 
sponding terms  will  also  form  a  proportion. 

307.  If  two  proportions  be  given  in  which  two  corresponding 
ratios  have  the  antecedent  of  one  equal  to  the  consequent  of 
the  other,  the  remaining  antecedent  and  consequent,  together 
with  the  products  of  the  corresponding  terms  of  the  other  two 
ratios,  will  form  a  proportion. 

308.  If  the  antecedents  of  a  proportion  are  equal,  the  con- 
sequents are  equal. 

308  (a).   Converse  of  308. 

309.  If  four  quantities  form  a  proportion,  their  like  powers 
and  like  roots  will  also  form  a  proportion. 

310.  If  three  quantities  form  a  proportion,  the  first  is  to 
the  third  as  the  square  of  the  first  is  to  the  square  of  the 
second. 

311.  If  four  quantities  form  a  proportion,  the  sum  of  the 
squares  of  the  first  two  terms  is  to  their  product  as  the  sum 
of  the  squares  of  the  last  two  is  to  their  product. 


68  PLANE  GEOMETRY. 

311  (a).   Substitute  "difference"  for  "sum"  in  311. 

312.  If  two  quantities  be  either  increased  or  diminished  by 
like  parts  of  each,  the  results  will  be  in  the  same  ratio  as  the 
quantities  themselves. 

312  (a).   If  three  terms  of  a  simple  proportion  are  equal 
respectively  to  the  three  corresponding  terms  of  another  pro- 
portion, the  fourth  terms  of  the  two  proportions  ^re  equal. 

PKOPOBTIOffAL  LINES. 

313.*  If  a  line  be  drawn  parallel  to  one  side  of  a  triangle, 
the  four  parts  into  which  it  divides  the  other  sides  will  form 

a  proportion. 

Post.  Let  ABC  be  any  tri- 
angle, and  S  T  a  line  drawn 
parallel  to  the  side  CB. 

We  are  to  prove  that  the 
four  parts  AS,  SB,  AT,  and 

TC  form  a  Pr°P°rtion- 
Dem.   Let  us  conceive  the 

line  DH  passing  through  the 
vertex  A  and  parallel  to  BC, 
to  move  toward  BC  and  remaining  always  parallel  to  it.  The 
instant  it  starts  there  will,  of  course,  be  two  points  of  inter- 
section, as  0  and  P.  It  is  evident  that  when  the  point  0  has 
reached  the  point  B,  the  point  P  will  have  reached  the  point  C. 
Why? 

Again,  if  Q  and  R  be  the  middle  points  of  the  sides  AB  and 
AC  respectively,  it  is  evident  that  when  the  point  0  reaches 
the  point  Q,  the  point  P  will  reach  the  point  R.  Why  ? 

Hence  when  the  point  O  has  moved  over  one-half  of  AB, 
the  point  P  has  moved  over  one-half  of  AC.  Similarly,  when 
0  has  moved  over  one-fourth,  one-tenth,  one-thousandth,  or 

*  See  Appendix. 


PLANE  GEOMETRY.  69 

one-7ith  of  AB,  P  has  moved  over  the  same  part  of  AC. 
Hence,  no  matter  what  the  position  of  the  moving  line,  AO  is 
always  the  same  part  of  AB  that  AP  is  of  AC;  that  is,  the 
ratio  of  AO  to  AB  is  the  same  as  the  ratio  of  AP  to  AC',  or 

AO  :  AB  :  :  AP  :  AC. 

Now,  since  by  Hyp.  ST  is  parallel  to  BC  when  point  0 
reaches  $,  point  P  reaches  T. 

Hence  AS  :  AB  : :  AT :  AC. 

But  AB-AS:AS::AC-AT:AT.  Why? 

But         AB  -AS  =  SB    and    AC-AT=TC. 

.'.  SB:AS::TC:  AT.  Why  ? 

Q.E.D. 

314.  If  a  line  be  draw'n  parallel  to  one  side  of  a  triangle, 
the  other  two  sides  and  either   pair  of  corresponding  parts 
form  a  proportion. 

Sug.  Use  result  obtained  in  previous  theorem,  and  consult  297. 

315.  Converse  of  313.  4 
Post.   Let  ACH  be  a  tri- 
angle, with  line  BD  drawn 

so  that 

AB:BC::AD:  DH. 


We  are  to  prove  BD  par- 
allel to  CH. 

Dem.  First,  suppose  BKto  be  drawn  through  B  parallel  to  CH. 

Then                         AB  :  BC  ::  AK:  KH.  Why? 

/.  AD:DH::  AK :  KH.  Why  ? 

.-.  AD  +DH:DH:  :  AK+KH :  KH;  Why  ? 
or                                  AH:DH::AH:KH. 

.-.  DH=KH.  Why? 

.-.  points  D  and  K  must  coincide.  Why  ? 


70  PLANE  GEOMETRY. 

What,  then,  must  be  the  relative  position  of  the  lines  BK 
and  BD  ?  Why  ? 

But  BK  was  drawn  parallel  to  CH.  Consequently  _BZ>, 
which  coincides  with  BK,  must  also  be  parallel  to  CPL  Q.E.D. 

316.  If  a  series  of  parallel  transversals,  intersecting  any 
two  straight  lines,  intercept  equal  distances  on  one  of  these 
lines,  they  will  also  intercept  equal  distances  on  the  other. 

317.  If  a  series  of  parallel  transversals,  intersecting  any 
number  of  straight  lines,  intercept  equal  distances  on  one  of 
these  lines,  they  will  also  intercept  equal  distances  on  each  of 
the  others. 

318.  If  two  lines  be  drawn  parallel  to  one  side  of  a  triangle 
intersecting  the  other  two  sides,  the  parts  thus  intercepted 
form  a  proportion. 

318  (a) .  If  a  line  be  drawn  parallel  to  one  side  of  a  triangle, 
this  side,  its  parallel,  and  either  of  the  other  two  sides,  together 
with  the  segment  of  the  latter  joining  the  third  side,  will  form 
a  proportion. 

319.  If  any  number  of  straight  lines  be  intersected  by  a 
series  of  parallel  transversals,  the  parts  thus  intercepted  by 
the  latter  form  a  proportion. 

320.  If  a  line  be  drawn  parallel  to  the  base  of  a  triangle, 
and  another  from  vertex  to  base,  the  parts  of  both  the  base 
and  its  parallel  will  form  a  proportion. 

321.  If  any  number  of  straight  lines  which  pass  through  a 
common  point  be  intersected  by  a  series  of  parallel  transversals, 
the  parts  intercepted  by  both  parallels  and  non-parallels  form 
a  proportion. 

322.  Converse  of  321. 

323.  The  bisector  of  an  angle  of  a  triangle  divides  the  oppo- 
site side  into  segments  proportional  to  the  other  two  sides. 

Sug.   From  the  vertex  of  the  bisected  angle  extend  one  of 


PLANE  GEOMETRY.  71 

the  sides,  and  from  one  of  the  other  vertices  draw  a  line 
parallel  to  the  bisector  meeting  the  side  extended. 

324.  If  the  bisector  of  an  exterior  angle  of  a  triangle  meet  one 
of  the  sides  extended,  this  side  plus  its  extension,  the  extension, 
and  the  other  two  sides  of  the  triangle,  form  a  proportion. 

Sug.  From  extremity  of  the  side  extended,  draw  a  parallel 
to  the  bisector. 

POLYGONS. 

325.  A  polygon  is  a  portion  of  a  plane  bounded  by  straight 
lines. 

What  is  the  least  number  of  sides  that  a  polygon  can  have  ? 
The  greatest  number  ? 

326.  The  word  "polygon  "  is  from  two  Greek  words  meaning 
many  angles. 

What  is  a  polygon  of  three  sides  usually  called  ?  It  is  also 
called  a  trlgon.  A  polygon  of  four  sides  is  usually  termed 
what  ?  It  is  also  called  a  tetragon.  A  polygon  of  five  sides 
is  called  a  pentagon,  one  of  six  sides  a  hexagon,  one  of  seven 
sides  a  heptagon,  one  of  eight  sides  an  octagon,  one  of  nine  sides 
a  nonagon  or  enneagon,  one  of  ten  sides  a  decagon,  one  of 
eleven  sides  an  undecagon,  one  of  twelve  sides  a  dodecagon,  and 
one  of  fifteen  sides  a  pentedecagon. 

327.  A  convex  polygon  is  one  each  of  whose  angles  is  less 
than  180°. 

A  concave  polygon  is  one  that  has  one  or  more  re-entrant 
angles ;  i.e.  where,  at  one  or  more  vertices,  the  angular  space 
within  the  polygon  is  more  than  180°. 

Whenever  polygons  are  mentioned  in  this  work,  convex 
polygons  are  meant  unless  otherwise  specified. 

The  diagonal  of  a  polygon  is  a  line  joining  any  two  vertices 
not  contiguous. 

328.  A  regular  polygon  is  one  that  is  both  equilateral  and 
equiangular. 


72  PLANE  GEOMETRY. 

The  perimeter  of  a  polygon  is  the  sum  of  its  sides.  How 
many  angles  has  each  of  the  above-named  polygons  ?  How 
does  the  number  of  sides  compare,  in  each  case,  with  the  num- 
ber of  angles  ?  How  many  angles,  then,  has  a  polygon  of  n 
sides  ?  How  many  diagonals  can  be  drawn  from  a  single 
vertex  in  each  of  the  above-named  polygons  ?  Compare  the 
number  of  diagonals  in  each  case  with  the  number  of  sides. 
If  the  polygon  has  n  sides,  how  many  diagonals  can  be  drawn 
from  a  single  vertex?  How  many  triangles  are  formed  by 
drawing  diagonals  from  a  single  vertex  in  each  of  the  above- 
named  polygons  ?  How  does  the  number  of  triangles  compare 
with  the  number  of  sides  ?  If  the  polygon  has  n  sides,  how 
many  triangles  would  be  formed  by  diagonals  similarly 
drawn  ? 

If  from  any  point  selected  at  random  in  a  polygon  lines  be 
drawn  to  the  vertices,  into  how  many  triangles  will  it  be  thus 
divided  ? 

329.  The  sum  of  the  interior  angles  of  a  polygon  is  equal  to 
two  right  angles,  taken  as  many  times,  less  two,  as  the  polygon 
has  sides ;  or  twice  as  many  right  angles  as  the  polygon  has 
sides,  less  four  right  angles. 

Sug.  Divide  the  polygon  up  into  triangles,  and  let  n  equal 
the  number  of  sides.  Having  found  an  expression  in  terms  of 
n,  and  a  right  angle  for  the  sum  of  the  angles,  what  would  be 
the  value  of  one  angle  if  the  polygon  were  equiangular? 

330.  The  sum  of  the  exterior  angles  of  any  polygon  formed 
by  extending  each  of  its  sides  in  succession  similarly,  is  equal 
to  four  right  angles. 

Sug.   Consult  Sect.  25  and  Theorem  325. 

PROBLEMS  OF  COMPUTATION. 

331.  Find  the  value  in  right  angles  of  the  sum  of  the  interior 
angles  of  a  pentagon,  hexagon,  octagon,  decagon,  dodecagon,  pen- 
tedecagon,  and  a  polygon  of  fifty-two  sides. 


PLANE  GEOMETRY.  73 

332.  Find  the  value,  in  terms  of  a  right  angle  as  the  unit, 
of  one  of  the  angles  of  an  equiangular  pentagon,  octagon,  dodec- 
agon, a  polygon  of  twenty  sides,  one  hundred  sides. 

333.  Find  the  values  of  the  above  angles  in  degrees,  also  in 
grades. 

334.  How  many  sides  has  an  equiangular  polygon,  the  sum 
of  four  angles  of  which  is  equal  to  seven  right  angles  ? 

335.  How  many  sides  has  an  equiangular  polygon,  the  sum 
of  three  angles  of  which  is  equal  to  five  right  angles  ? 

336.  How  many  sides  has  an  equiangular  polygon,  the  sum 
of  nine  angles  of  which  is  equal  to  sixteen  right  angles  ? 

337.  How  many  sides  has  the  polygon,  the  sum  of  whose 
interior  angles  is  equal  to  the  sum  of  its  exterior  angles  ? 

338.  How  many  sides  has  the  polygon,  the  sum  of  whose 
interior  angles  is  double  that  of  its  exterior  angles  ? 

339.  How  many  sides  has  the  polygon,  the  sum  of  whose 
exterior  angles  is  double  that  of  its  interior  angles  ? 

340.  How  many  sides  has  the  polygon,  the  sum  of  whose 
interior  angles  is  equal  to  nine  times  the  sum  of  its  exterior 
angles  ? 

341.  How  many  sides  has  the  equiangular  polygon,  when 
the  sum  of  nine  of  its  interior  angles  is  four  times  the  sum  of 
its  exterior  angles  ? 

342.  How  many  sides  has  the  equiangular  polygon,  when 
the  sum  of  five  of  its  interior  angles  is  equal  to  two  and  one- 
fourth  times  the  sum  of  its  exterior  angles  ? 

343.  How  many  sides  has  the  equiangular  polygon  when 

(a)  5^==9frt.z£?      (e)  8  A  =  3J  times  its  ext.  A  ? 

(b)  4  «  =  7J    "  (/)  6  "  =  3f     "        "        " 

(c)  16  "  =  31     «  (g)  7  "  =  6|rt.  A  ? 

(d)  5  «  =  8       «  (h)  3  "  =  5|     « 


74  PLANE  GEOMETRY. 

SIMILAR  FIGUKES. 

344.  Similarity  in  general  means  likeness  of  form;  i.e.  two 
figures  are  said  to  be  similar  when  they  have  the  same  shape, 
although  they  may  differ  in  size. 

Similar  geometrical  figures  are  those  whose  homologous  angles 
are  equal,  and  whose  homologous  sides  form  a  proportion. 

345.  Homologous  sides  of  similar  polygons  are  those  which 
join  the  vertices  of  equal  angles  in  the  respective  polygons. 

In  similar  triangles  the  homologous  sides  are  those  that  are 
opposite  the  equal  angles. 

The  pupil  should  be  careful  to  observe  at  the  outset  that 
similarity  involves  two  things ;  viz.  equality  of  angles  and  pro- 
portionality of  sides. 

346.  Two  polygons  are  said  to  be  mutually  equiangular  when 
the  angles  of  one  are  equal  respectively  to  the  corresponding 
angles  of  the  other,  and  mutually  equilateral  when  each  side  of 
one  has  an  equal  side  in  the  other. 

SIMILAE  TEIAttaLES. 

347.  If  two  triangles  be  mutually  equiangular,  they  are 
similar. 

A 


p 

Post.   Let  ABH  and  KDN  be  two  triangles  having 

and 


We  are  to  prove  the  triangles  similar  ;  i.e.  that  their  homol- 
ogous sides  form  a  proportion. 


PLANE  GEOMETRY.  75 

Cons.  Mark  off  on  HB  a  distance  HP  equal  to  KN,  and  on 
HA  a  distance  HQ  equal  to  DN,  and  join  PQ. 

Dem.  What  relation  exists  between  the  two  A  RDN  and 
PQH?  Why? 

What,  then,  must  be  the  relation  between  the  A  PQH  and 
D  ?  What  between  A  QPH  and  K? 

What,  then,  must  be  the  relation  between  the  A  PQH  and 
A  ?  What  between  A  QPH  and  B  ?  Why  ? 

What,  then,  must  be  the  relative  position  of  the  lines  AB 
and  QP  ?  Why  ? 

Apply  Theorem  314,  and  the  remainder  of  the  demonstration 
should  be  easy.  Q.E.D. 

348.  If  two  triangles  have  two  angles  of  one  equal  respec- 
tively to  two  angles  of  the  other,  the  two  triangles  are  similar. 

349.  If  two  right  triangles  have  an  acute  angle  of  one  equal 
to  an  acute  angle  of  the  other,  the  two  triangles  are  similar. 

350.  If  two  triangles  have  an  angle  in  each  equal,  and  the 
sides  including  those  angles  form  a  proportion,  the  two  tri- 
angles are  similar. 

Sug.  Apply  one  to  the  other  so  that  the  equal  angles  shall 
coincide ;  then  consult  Theorem  315.  Or  proceed  in  a  manner 
similar  to  that  in  Theorem  347. 

351.  If  two  triangles  have  their  sides  respectively  parallel, 
they  are  similar. 

Sug.  Consult  94  and  98,  then  three  possible  hypotheses 
regarding  the  relations  of  the  respective  pairs  of  angles. 

352.  If  two  triangles  have  their  sides  respectively  perpen- 
dicular to  each  other,  they  are  similar. 

Sug.  Consult  95  and  98,  and  see  suggestions  to  351.  Consult 
the  caution  in  Sect.  55. 

353.  If  two  triangles  be  similar,  their  homologous  altitudes 
are  in  the   same   ratio   as   either  pair  of  homologous   sides 
including  the  vertical  angle. 


76  PLANE  GEOMETRY. 

354.  If  two  triangles  be  similar,  their  homologous  altitudes 
are  in  the  same  ratio  as  their  homologous  bases. 

355.  If  the  homologous  sides  of  two  triangles  form  a  pro- 
portion, the  triangles  are  similar. 


Post.  Let  ABC  and  DHK  be  two  triangles,  their  homol- 
ogous sides  forming  the  continued  proportion. 

AB  :  DHi:  AC  :  DK : :  BC  :  HK. 

We  are  to  prove  that  the  two  triangles  are  similar ;  i.e.  that 
they  are  mutually  equiangular. 

Cons.  Make  NC  equal  to  DK,  PC  equal  to  HK,  and  join 
NP. 

Dem.  In  the  given  proportion  substitute  for  DK  and  HK 
their  equals  NC  and  PC.  Then 

AC:NC::BC:  PC. 

What  is  true,  then,  of  the  two  triangles  ABC  and  NPC  ? 
See  Theorem  350. 

.-.  AB:NP::BC:  PC.  Why  ? 

But  by  Hyp., 

AB-.DH'.'.BC'.HK. 

What  is  true  of  the  first  terms  of  these  two  proportions  ? 
Of  the  third  terms  ?  Of  the  last  terms  ?  What  must  be  true, 
then,  of  the  second  terms  ? 

The  pupil  should  finish  the  demonstration  without  difficulty. 


PLANE  GEOMETRY.  77 

356.  If  two   polygons    be    similar,    the    diagonals    drawn 
from  homologous   vertices   will   divide   them   into   the  same 
number  of  triangles,    similar    two    and    two,   and   similarly 
placed. 

357.  Converse  of  356. 

358.  The  perimeters  of  two  similar  polygons  are  in  the  same 
ratio  as  any  two  homologous  sides  or  any  two  homologous 
diagonals. 

359.  The  perimeters  of  two  similar  polygons  are  in  the  same 
ratio  as  the  bisectors  of  any  two  homologous  angles  or  any 
two  homologous  lines  howsoever  drawn. 

360.  If,  in  a  right  triangle,  a  perpendicular  be  drawn  to  the 
hypothenuse  from  the  vertex  of  the  right  angle, 

I.  the  two  triangles  thus  formed  are  similar  to  the  original 
triangle  and  to  each  other. 

II.  the  perpendicular  and  the  two  segments  of  the  hypothe- 
nuse form  a  proportion. 

(Where  three  quantities  form  a  proportion,  what  relation 
must  one  of  these  quantities  sustain  to  the  other  two  ?) 

III.  either  leg,  the  hypothenuse,  and  that  segment  of  the 
latter  which  joins  the  former,  form  a  proportion. 

(The  pupil  should  preserve  these  proportions  for  use  further 
on.) 

IV.  the  squares  of  the  two  legs  and  the  two  segments  of 
the  hypothenuse  form  a  proportion. 

Sug.   Use  the  two  proportions  in  III.  by  Theorem  293,  and 
divide. 

V.  the  square  of  the  hypothenuse  bears  the  same  ratio  to 
the  square  on  either  leg,  as  the  hypothenuse  bears  to  that  seg- 
ment of  the  latter  joining  that  leg. 

VI.  the  two  legs,  the  hypothenuse,  and  the  perpendicular 
form  a  proportion. 


78 


PLANE  GEOMETRY. 


360  (a).  The  square  of  the  hypothemise  of  a  right  triangle 
is  equal  to  the  sum  of  the  squares  of  the  two  legs.  See 
Theorem  426. 

Sug.  Use  result  in  360,  IV.,  by  composition,  and  then  com- 
pare with  360,  V. 

361.  The  square  of  the  diagonal  of  a  square  is  equal  to  twice 
the  square  of  one  of  its  sides. 


PKOJEOTION. 

362.  The  projection  of  one  line  upon  another  is  that  portion 
of  the  latter  included  between  two  perpendiculars  to  the  latter 
drawn  from  the  extremities  of  the  former. 


K'  B 


Fig.  I. 


For  example,  HKis  the  projection  of  the  line  CD  upon  AB, 
and  RS  is  the  projection  of  RQ  on  NP;  it  is  also  the  projec- 
tion of  RQ  on  RF.  Mention  all  the  cases  of  projection  in 
Figure  III.,  the  dotted  lines  being  perpendicular  to  the  respec- 
tive sides. 

363.  In  any  triangle  the  square  of  the  side  opposite  an  acute 
angle  is  equal  to  the  sum  of  the  squares  of  the  other  two  sides, 
minus  twice  the  product  of  one  of  those  sides  and  the  projec- 
tion of  the  other  upon  that  side. 

If  C  be  the  acute  angle,  then  by  a  glance  at  the  following 
diagram  it  will  be  readily  seen  that  there  may  be  two  cases 
depending  upon  whether  the  projection  involves  an  extensiot 


PLANE  GEOMETRY. 


79 


of  one  side,  which  will  evidently  be  the  case  if  the  side  to  be 
projected  is  opposite  an  obtuse  angle. 


C      V 


Fig.  I. 

We  are  to  prove 


In  Case  I. 
In  Case  II. 


Fig.  H. 


=BU2  +AU*  -  2BC  -  DO. 
DB  =BC  -DC. 
DB=DC-BC. 


The  square  of  either  of  these  two  equations  gives  the  same 
result  ;  hence  to  the  square  add  the  squares  of  the  perpendicu- 
lar, and  then  combine  the  terms  by  using  Theorem  360  (a). 

364.  In  an  obtuse  triangle  the  square  of  the  side  opposite 
the  obtuse  angle  is  equal  to  the  sum  of  the  squares  of  the  other 
two  sides,  plus  twice  the  product  of  one  of  those  sides  and  the 
projection  of  the  other  upon  that  side. 

Sug.  Form  an  equation  by  placing  the  projection  of  the  side 
opposite  the  obtuse  angle  equal  to  the  sum  of  its  two  parts, 
then  proceed  as  in  363. 

365.  If  any  median  of  a  triangle  be  drawn, 

I.  the  sum  of  the  squares  of  the  other  two  sides  is  equal  to 
twice  the  square  of  one-half  the  bisected  side,  plus  twice  the 
square  of  the  median  ;  and 

II.  the  difference  of  the  squares  of  the  other  two  sides  is 
equal  to  twice  the  product  of  the  bisected  side  and  the  pro- 
jection of  the  median  upon  that  side. 

Sug.    Use  363  and  364,  then  combine  the  resulting  equations. 


80  PLANE  GEOMETRY. 

366.  In  any  quadrilateral  the  sum  of  the  squares  of  the  four 
sides  is  equal  to  the  sum  of  the  squares  of  the  diagonals,  plus 
four  times  the  square  of  the  line  joining  the  middle  points  of 
the  diagonals. 

Sug.  Use  365.  What  modification  of  the  above  theorem 
would  result  if  the  quadrilateral  were  a  parallelogram  ? 

367.  If  from  any  point,  selected  at  random,  in  the  circum- 
ference of  a  circle,  a  perpendicular  be  drawn  to  a  diameter, 
this  perpendicular  will  be  a  mean  proportional  between  the 
two  segments  of  the  diameter. 

368.  If  two  chords  of  a  circle  intersect  each  other,  the  four 
parts  form  a  proportion. 

369.  If  two  secants  be  drawn  from  the  same  point  without 
the  circle,  the  entire  secants  and  the  parts  that  are  without 
the  circle  form  a  proportion. 

370.  If,  from  the  same  point,  a  tangent  and  secant  to  a  cir- 
cle be  drawn,  the  tangent,  the  secant,  and  that  part  of  the 
latter  that  is  outside  the  circle,  will  form  a  proportion. 

371.  The  two  tangents  to  two  intersecting  circles  from  any 
point  in  their  common  secant  are  equal. 

Sug.   Consult  370. 

372.  If  two  circles  intersect  each  other,  their  common  chord 
extended  bisects  their  common  tangents. 

ADVANCE  THEOREMS. 

373.  In  any  triangle  the  product  of  any  two  sides  is  equal 
to  the  diameter  of  the  circumscribed  circle  multiplied  by  the 
perpendicular  drawn  to  the  third  side  from  the  vertex  of  the 
angle  opposite. 

Sug.  Construct  the  diameter  from  the  same  vertex  as  the 
perpendicular,  and  join  its  extremity  with  one  of  the  other 
vertices,  making  a  right  triangle  similar  to  the  one  formed  by 
the  perpendicular,  whence  the  necessary  proportion. 


PLANE  GEOMETRY.  81 

374.  In  any  triangle  the  product  of  any  two  sides  is  equal 
to  the  product  of  the  segments  of  the  third  side,  formed  by 
the  bisector  of  the  opposite  angle,  plus  the  square  of  the 
bisector. 

Sug.  Circumscribe  a  circle,  and  extend  the  bisector  to  the 
circumference,  and  connect  its  extremity  with  one  of  the  other 
vertices  of  the  triangle.  Then  if  the  vertices  of  the  triangle 
be  lettered  A,  B,  and  (7,  and  the  bisector  be  drawn  from  A, 
and  D  the  point  where  the  bisector  crosses  the  side  BC,  and 
E  the  extremity  of  the  bisector  extended,  then  the  triangles 
BAD  and  ACE  can  be  proved  similar. 

375.  If  one  side  of  a  right  triangle  is  double  the  other,  the 
perpendicular,  from  the  vertex  of  the  right  angle  to  the  hypoth- 
enuse,  divides  it  into  segments  which  are  to  each  other  as  1 
to  4. 

376.  A  line  parallel  to  the  bases  of  a  trapezoid,  passing 
through  the  intersection  of  the  diagonals,  and  terminating  in 
the  non-parallel  sides,  is  bisected  by  the  diagonals. 

377.  In  any  triangle  the  product  of  any  two  sides  is  equal 
to  the  product  of  the  segments  of  the  third  side,  formed  by  the 
bisector  of  the  exterior  angle  at  the  opposite  vertex,  minus  the 
square  of  the  bisector. 

Sug.   Consult  Theorem  374. 

378.  The  perpendicular,  from  the  intersection  of  the  medians 
of  a  triangle,  upon  any  straight  line  in  the  plane  of  the  tri- 
angle, is  one-third  the  sum  of  the  perpendiculars  from  the  ver- 
tices of  the  triangle  upon  the  same  line. 

379.  If  two  circles  are  tangent  to  each  other,  their  common 
tangent  and  their  diameters  form  a  proportion. 

380.  If  two  circles  are  tangent  internally,  all  chords  of  the 
greater  circle  drawn  from  the  point  of  contact  are  divided  pro- 
portionally by  the  circumference  of  the  smaller. 


82  PLANE  GEOMETRY. 

381.  In  any  quadrilateral  inscribed  in  a  circle,  the  product 
of  the  diagonals  is  equal  to  the  sum  of  the  products  of  the 
opposite  sides. 

Sug.  From  one  vertex  draw  a  line  to  the  opposite  diagonal, 
making  the  angle  formed  by  it  and  one  side  equal  to  the  angle 
formed  by  the  other  diagonal  and  side  which  meets  the  former. 

382.  If  three  circles  whose  centres  are  not  in  the   same 
straight  line  intersect  one  another,  the  common  chords  will 
intersect  each  other  at  one  point. 

383.  If  two  chords  be  perpendicular  to  each  other,  the  sum 
of  the  squares  of  the  four  segments  is  equal  to  the  square  of 
the  diameter. 

384.  The  sum  of  the  squares  of  the  diagonals  of  a  quadri- 
lateral is  equal  to  twice  the  sum  of  the  squares  of  the  lines 
joining  the  middle  points  of  the  opposite  sides. 

PROBLEMS  OF  COMPUTATION. 

385.  The  chord  of  one-half  a  certain  arc  is  9  inches,  and  the 
distance  from  the  middle  point  of  this  arc  to  the  middle  of 
its  subtending  chord  is  3  inches.     Find  the  diameter  of  the 
circle. 

386.  The  external  segments  of  two  secants  to  a  circle  from 
the  same  point  are  10  inches  and  6  inches,  while  the  internal 
segment  of  the  former  is  5  inches.     What  is  the  internal  seg- 
ment of  the  latter  ? 

387.  The  hypothenuse  of  a  right  triangle  is  16  feet,  and  the 
perpendicular  to  it  from  the  vertex  of  the  opposite  angle  is 
5  feet.     Find  the  values  of  the  legs  and  the  segments  of  the 
hypothenuse. 

388.  The  sides  of  a  certain  triangle  are  6,  7,  and  8  feet 
respectively.     In  a  similar  triangle  the  side  corresponding  to 
8  is  40.     Find  the  other  two  sides. 


PLANE  GEOMETRY.  83 

389.  The  sides  of  a  certain  triangle  are  9,  12,  and  15  feet 
respectively.     Find  the  segments  of  the  sides  made  by  the 
bisectors  of  the  several  angles. 

390.  If  a  vertical  rod  6  feet  high  cast  a  shadow  4  feet  long, 
how  high  is  the  tree  which,  at  the  same  time  and  place,  casts 
a  shadow  90  feet  long  ? 

391.  The  perimeters  of  two  similar  polygons  are  200  and 
300  feet  respectively,  and  one  side  of  the  former  is  24  feet. 
What  is  the  corresponding  side  of  the  latter  ? 

392.  How  long  must  a  ladder  be  to  reach  a  window  24  feet 
high,  if  the  lower  end  of  the  ladder  is  10  feet  from  the  side  of 
the  house  ? 

393.  Find  the  lengths  of  the  longest  and  the  shortest  chord 
that  can  be  drawn  through  a  point  6  inches  from  the  centre  of 
a  circle  whose  radius  is  10  inches. 

394.  The  distance  from  the  centre  of  a  circle  to  a  chord  10 
inches  long  is  12  inches.     Find  the  distance  from  the  centre 
to  a  chord  24  inches  long. 

395.  The  radius  of  a  circle  is  5  inches.     Through  a  point 
3  inches  from  the  centre  a  diameter  is  drawn,  and  also  a  chord 
perpendicular  to  the  diameter.     Find  the  length  of  this  chord, 
and  the  distance  from  one  end  of  the  chord  to  the  ends  of  the 
diameter. 

396.  Through  a  point  10  feet  from  the  centre  of  a  circle 
whose  radius  is  6  feet  tangents  are  drawn.     Find  the  lengths 
of  the  tangents  and  of  the  chord  joining  the  points  of  contact. 

* 

397.  If  a  chord  8  feet  long  be  3  feet  from  the  centre  of  the 
circle,  find  the  radius  and  the  distances  from  the  end  of  the 
chord  to  the  ends  of  the  diameter  which  bisects  the  chord. 

398.  Through  a  point  5  inches  from  the  centre  of  a  circle 
whose  radius  is  13  inches  any  chord  is  drawn.     What  is  the 


84  PLANE  GEOMETRY. 

product  of  the  two  segments  of  the  chord  ?    What  is  the  length 
of  the  shortest  chord  that  can  be  drawn  through  that  point  ? 

399.  From  the  end  of  a  tangent  20  inches  long  a  secant  is 
drawn  through  the  centre  of  a  circle.     If  the  exterior  segment 
of  this  secant  be  8  inches,  what  is  the  radius  of  the  circle  ? 

400.  A  tangent  12  feet  long  is  drawn  to  a  circle  whose 
radius  is  9  feet.     Find  the  external  segment  of  a  secant  through 
the  centre  from  the  extremity  of  the  tangent. 

401.  The  span  of  a  roof  is  28  feet,  and  each  of  its  slopes 
measures  17  feet  from  the  ridge  to  the  eaves.     Find  the  height 
of  the  ridge  above  the  eaves. 

402.  A  ladder  40  feet  long  is  placed  so  as  to  reach  a  window 
24  feet  high  on  one  side  of  the  street,  and  on  turning  the 
ladder  over  to  the  other  side  of  the  street,  it  just  reaches  a 
window  32  feet  high.     What  is  the  width  of  the  street  ? 

403.  The  bottom  of  a  ladder  is  placed  at  a  point  14  feet 
from  a  house,  while  its  top  rests  against  the  house  48  feet 
from  the  ground.     On  turning  the  ladder  over  to  the  other 
side  of  the  street  its  top  rests  40  feet  from  the  ground.     Find 
the  width  of  the  street. 

404.  One  leg  of  a  right  triangle  is  3925  feet,  and  the  differ- 
ence between  the  hypothenuse  and  other  leg  is  625  feet.     Find 
the  hypothenuse  and  the  other  leg. 

AEEAS, 

405.  The  area  of  a  surface  is  its  numerical  measure ;  i.e. 
the  numerical  expression  for  the  number  of  times  it  contains 
another  surface  arbitrarily  assumed  as  a  unit  of  measure.    For 
example,  the  area  of  the  floor  of  a  room  is  the  number  of 
times  it  contains  some  one  of  the  common  units  of  surface, 
square  foot,  square  yard,  etc.     This  unit  of  surface  is  called 
the  superficial  unit.     The  most  convenient  superficial  unit  is 


PLANE  GEOMETRY. 


85 


the  square  of  the  linear  unit.  The  square  of  a  linear  unit  (or 
line)  is  the  area  of  the  square  constructed  with  that  linear 
unit  for  its  sides.  Surfaces  that  are  equal  in  area  are  said  to 
be  equivalent. 

406.*   If  two  rectangles  have  equal  altitudes,  their  areas 
will  be  in  the  same  ratio  as  their  bases. 

CASE  I.  —  When  the  bases  are  commensurable  (262). 


D 


w  w  w" 


-JV" 


r 


Q 


s 

s' 

1 
J 

J 

? 

K 

t    1 

V    t" 

a  a'  a" 

Post.  Let  ABCD  and  HKNP  be  two  rectangles,  having  their 
altitudes  AD  and  HP  equal,  and  their  bases  AB  and  HK  com- 
mensurable. Designate  their  areas  by  Q  and  R  respectively. 

We  are  to  prove        Q:R::AB:  HK. 

Dem.  Since  AB  and  HK  are  commensurable,  they  must 
have  a  common  measure  (262). 

Suppose  this  common  measure  to  be  contained  in  AB  n  times, 
and  in  HK  m  times. 

At  the  several  points  of  division,  a,  a',  a",  etc.,  t,  t1,  t",  etc., 
construct  ab,  a'b',  a"b",  etc.,  perpendicular  to  AB,  and  tw} 
t'w',  t"w",  etc.,  perpendicular  to  HK. 

Then  the  rectangle  ABCD  will  be  divided  into  n  equal 
rectangles  r,  r',  r",  etc.,  and  the  rectangle  HKNP  will  be 
divided  into  m  equal  rectangles  s,  s',  s",  etc.  These  smaller 
rectangles  are  also  equal  to  one  another. 

(The  pupil  should  demonstrate  the  above  propositions.) 

Hence  AB  :  HK : :    n    :    m,      (264.) 

and  Q   :   R    ::    n    :    m.       (264.) 

.-.    Q   :  R    ::AB:HK.     (288.)  Q.E.D. 


*  See  Appendix. 


86 


PLANE  GEOMETRY. 


CASE  II.  — When  the  bases  are  incommensurable. 

Suppose  AB  to  be  divided  into  any  number  of  equal  parts, 
as  «,  and  that  one  of  these  parts  be  applied  to  HK  and  found 
to  be  contained  m  times  with  a  remainder  less  than  that  part. 
From  the  points  of  division  construct  lines  as  in  Case  I. 
Then  the  rectangle  ABCD  will  be  divided  into  n  equal  rec- 
tangles, and  rectangle  KNPH  into  m  equal  rectangles,  with  a 
remaining  rectangle  less  than  one  of  the  m  rectangles. 

Then  Q :  R  and  AB  :  HK  are  two  incommensurable  ratios. 

T-»  TJ  -J 

(1)       5:>S     and       -   <-  +  -. 
Q        n  Q        n      n 


and 
AB       n  AB       n      n 

Hence  the  two  ratios  R  :  Q  and  HK :  AB  have  the  same 
approximate  numerical  value,  whatever  the  magnitude  of  n. 

.'.     R:Q::HK:AB,     (289.) 
or  Q'.R'.-.AB:  HK.  Q.E.D. 

407.  If  two  rectangles  have  equal  bases,  their  areas  are  in 
the  same  ratio  as  their  altitudes. 

Sug.   Consider  their  altitudes  as  bases,  and  bases  as  alti- 
tudes, and  proceed  as  in  406. 

408.  The  areas  of  any  two  rectangles  have  the  same  ratio 
as  the  products  of  their  respective  bases  and  altitudes. 

A  rt          A  B  n 


A. 

G         D 
K 

C 

"*, 

N 
P 

H 

K 

TT 

a 

Kg.  I. 


Fig.  H. 


PLANE  GEOMETRY.  87 

(By  the  product  of  two  lines  is  meant  the  product  of  the 
numbers  which  represent  them  when  both  are  measured  by 
the  same  linear  unit.) 

Post.  Let  ABCD  and  HKNP  be  any  two  rectangles,  with 
DC  and  PN  their  respective  bases.  Let  us  designate  their 
areas  by  A  and  a  respectively,  their  altitudes  by  H  and  h, 
and  their  bases  by  B  and  b. 

We  are  to  prove  that  A  :  a  :  :  B  x  H  :  b  X  h. 

Dem.  Consider  the  two  rectangles  so  placed  that  the  ver- 
tices C  and  K  shall  coincide,  and  the  four  angles  all  equal. 

Then  BCH  is  one  straight  line,  as  is  also  DKN.     Why  ? 

Complete  the  rectangle  BQNC.  Then  considering  CN  the 
base  of  rectangle  BQNC,  it  has  the  same  altitude  as  rectangle 
ABCD. 

.-.  Area  ABCD  :  Area  BQNC  :  :  DC  :  KN.     (1)       Why  ? 

Again,  considering  BC  the  base  of  the  rectangle  BQNC, 

Area  HKNP  :  Area  BQNC  :  :  HK  :  BC,      (2) 
or         Area  BQNC  :  Area  HKNP  :  :  BC  :  HK     (3)      Why  ? 
Hence 

Area  ABCD  :  AreafflOTP  :  :  BC  X  DC  :  HKxKN, 

(See  Theorem  307.) 
or  A:  a::BxH  :b  X  h.  Q.E.D. 

408  (a).  Hence  the  above  demonstration  shows  that  if  the 
bases  and  altitudes  of  any  two  rectangles  be  measured  by  the 
same  linear  unit,  the  ratio  of  their  respective  products  (see 
408)  will  be  the  ratio  of  their  respective  surfaces,  and  conse- 
quently either  may  be  assumed  as  the  measure  of  the  other. 

For  example,  suppose  the  linear  unit  to  be  contained  5  times 
in  AD,  18  times  in  DO,  3  times  in  HP,  and  6  times  in  PN. 

Then  A  :  a  :  :  5  X  18  :  3  X  6, 

A     90 


or 

a 


88 


PLANE  GEOMETRY. 


This  simply  means  that,  using  a  as  the  unit,  the  area  A  is 
5  times  as  large  as  the  area  a ;  or  that  the  area  a,  using  A  as 
the  unit  surface,  is  ^  as  large  as  A. 

Practically  it  is  more  convenient  to  compare  the  areas  of 
both  rectangles  with  the  square  of  the  linear  unit  (405)  as  a 
unit  of  surface.  This  comparison  is  formally  shown  in  the 
enunciation  and  demonstration  of  the  following  theorem. 

409.  The  area  of  a  rectangle  is  measured  by  the  product  of 
its  base  and  altitude. 


Post.  Let  ABCD  be  any  rectangle ;  and  in  whatever  linear 
unit  the  base  and  altitude  be  expressed,  let  r  be  a  square  whose 
sides  are  the  same  unit.  Designate  its  area  by  R,  and  its  base 
and  altitude  by  6  and  h  respectively. 


We  are  to  prove 
Dem. 


1x1,      (408.) 


or 


R      b  x  h  . 
r       lxl? 

or,  since  r  is  the  unit  of  area, 

R  =  bxh. 


Q.E.D. 


When  the  base  and  altitude  are  commensurable,  this  is  ren- 
dered evident  by  dividing  the  rectangle 
into  squares,  each  equal  to  the  superficial 
unit,  as  shown  in  annexed  figure.  The 
above  demonstration,  however,  includes  the 
case  when  the  base  and  altitude  are  in- 
commensurable. 


PLANE  GEOMETRY. 


89 


410.  If  a  parallelogram  and  rectangle  have  the  same  or  equal 
bases  and  the  same  or  equal  altitudes,  they  are  equivalent. 


K 


Sug.  Place  them  so  that  their  bases  shall  coincide,  as  repre- 
sented in  above  diagram.  Then  AB  and  HK  are  in  the  same 
straight  line.  Why  ? 

Prove  equality  of  the  triangles  ADH  and  BCK,  and  apply 
Axiom  VII. 

411.  The  area  of  a  parallelogram  is  measured  by  the  prod- 
uct of  its  base  and  altitude. 

412.  If  two  parallelograms  have  the  same  or  equal  bases 
and  the  same  or  equal  altitudes,  they  are  equivalent. 

413.  If  two  parallelograms  have  the  same  or  equal  bases, 
their  areas  are  in  the  same  ratio  as  their  altitudes. 

414.  If  two  parallelograms  have  the  same  or  equal  altitudes, 
their  areas  are  in  the  same  ratio  as  their  bases. 

415.  The  areas  of  any  two  parallelograms  are  in  the  same 
ratio  as  the  products  of  their  respective  bases  and  altitudes. 

416.  If  a  triangle  and  parallelogram  have  the  same  or  equal 
bases  and  the  same  or  equal  altitudes,  the  area  of  the  latter 
is  double  that  of  the  former. 

Sug.   Place  them  so  that  their  bases  will  coincide. 

417.  The  area  of  a  triangle  is  measured  by  one-half  the 
product  of  its  base  and  altitude. 

418.  If  two  triangles  have  the  same  or  equal  bases  and  the 
same  or  equal  altitudes,  they  are  equivalent. 

419.  If  two  triangles  have  the  same  or  equal  bases,  their 
areas  are  in  the  same  ratio  as  their  altitudes. 


90  PLANE  GEOMETRY. 

420.  If  two  triangles  have  the  same  or  equal  altitudes,  their 
areas  are  in  the  same  ratio  as  their  bases. 

421.  The  areas  of  any  two  triangles  are  in  the  same  ratio  as 
the  products  of  their  respective  bases  and  altitudes. 

422.  The  area  of  a  trapezoid  is  equal  to  the  product  of  its 
altitude  and  one-half  the  sum  of  its  parallel  sides. 

423.  The   square  described  upon  the  sum  of  two  lines   is 
equivalent  to  the  sum  of  the  squares  upon  the  two  lines,  plus 
twice  the  rectangle  formed  by  the  two  lines. 

424.  The  square  described  on  the  difference  of  two  lines  is 
equivalent  to  the  sum  of  the  squares  upon  the  two  lines,  minus 
twice  the  rectangle  formed  by  the  two  lines. 

425.  The  rectangle  formed  by  the  sum  and  difference  of  two 
lines  is  equivalent  to  the  difference  of  the  squares  upon  the 
two  lines. 

426.  The  square  described  upon  the  hypothenuse  of  a  right 
triangle  is  equivalent  to  the  sum  of  the  squares  described 
upon  the  legs. 

This  theorem  was  first  demonstrated  by  Pythagoras,  about 
450  B.C.,  and  hence  is  called  the  Pythagorean  theorem.  It  has 
been  a  favorite  one  with  mathematicians,  and  consequently 
about  fifty  different  demonstrations  of  it  have  been  recorded. 
Among  the  many  diagrams  used,  the  following  are  a  few  ;  and 
it  is  hoped  that  the  pupil  will  endeavor  to  demonstrate  it  for 
himself,  using  either  one  of  these,  or,  preferably,  one  of  his 
own.  No.  I.  is  the  diagram  used  by  Euclid  in  the  demonstra- 
tion of  this  theorem,  constituting  his  famous  "47th."  No. 
VII.  is  the  diagram  used  by  the  late  President  Garfield,  who 
was  said  to  have  utilized  his  leisure  hours  in  Congress  in 
mathematical  investigation.  In  each  figure  ABC  is  the  given 
triangle,  and  in  No.  VII.  ADC  is  one-half  the  square  upon 
the  hypothenuse.  DH  is  drawn  parallel  to  BC  to  meet  BA 
extended.  The  method  is  algebraic,  and  involves  an  equation 
between  the  sum  of  the  areas  of  the  three  triangles  and  the 
trapezoid  BCDH. 


PLANE  GEOMETRY. 


91 


Fig.VH. 


Fig.  VI. 


92 


PLANE  GEOMETRY. 


Fig.  VIH. 

See  also  Theorem  360  (a). 


Fig.  IX. 


427.   The    following    demonstration  is  original    with    the 
author : 

A 


Post.  Let  AC  be  any  chord  in  the  circle  DAC,  and  DK  a 
diameter  perpendicular  to  AC.  Also  let  DH  be  drawn  inter- 
secting AC.  Join  HK-j  then 


but 
hence 

but 


DN  :  DK  :  :  DB  :  DH, 
.'.  DNx  DH=DKx  DB-, 

DK=DB+BK; 

DN  x  DH  =  DB  (DB  +  BK). 

DN  x  DH  =  DB~  +  DB  x  BK-, 

DB  x  BK=  AB  x  BO  = 
.-.  DN  X  Z>#= 


PLANE  GEOMETRY. 


93 


Now  conceive  chord  DH  to  revolve  about  D  as  a  centre 
until  the  point  H  coincides  with  point  A  ; 

then  DN  =  DH  =  DA. 

.-.  DA*=  DB2+  Aff.  Q.E.D. 

For  other  methods  of  demonstrating  this  theorem,  see  Jour- 
nal of  Education,  June  23  and  July  7,  1887  ;  and  May  24  and 
June  28,  1888. 

427  (a).   Converse  of  426. 

428.  The  diagonal  of  a  square  is  incommensurable  with  its 
side. 

Sug.  Find  value  of  diagonal  in  terms  of  its  side  by  Theorem 
426. 

429.  If  two  triangles  have  an  angle  in  each  equal,  their 
areas  are  in  the  same  ratio  as  the  products  of  the  sides  which 
include  the  equal  angles. 


Post.   Let  ABC  and  DHKke  two  triangles  with 


Designate  their  areas  by  S  and  s,  and  the  sides  opposite  the 
several  angles  by  the  corresponding  small  letters. 
We  are  to  prove  that 

S  :  s::cxb:7cx  d. 

Apply  the  triangles  so  that  the  equal  angles  shall  coincide, 
or  extend  the  sides  of  one  until  they  equal  the  corresponding 
sides  of  the  other,  as  in  the  above  diagram. 


94  PLANE  GEOMETRY. 

Dem.     I.  Area  DHK  :  Area  NHK: :  DH :  NH.          Why  ? 
II.  Area  NHK  :  Area  NHP  : :  HK :  HP.          Why  ? 
Whence 

AreaDffiT  :  Area  NHP  : :  DHx  HK  :  NH*  HP. 

Why? 
or  S  :s::kxd:cxb.  Q.E.D. 

430.  The  areas  of  two  similar  triangles  have  the  same  ratio 
as  the  squares  of  any  two  homologous  sides. 

Sug.  Consult  421  and  354. 

431.  The  areas  of  two  similar  polygons  are  in  the  same  ratio 
as  the  squares  of  any  two  homologous  sides. 

Sug.  Consult  356,  43Q,  and  303. 

432.  The  areas  of  two  similar  polygons  are  in  the  same  ratio 
as  the  squares  of  any  two  homologous  diagonals. 

433.  Any  two  homologous  sides  or  diagonals  of  two  similar 
polygons  are  in  the  same  ratio  as  the  square  roots  of  their 
areas. 

434.  If  similar  polygons  are  described  upon  the  sides  of  a 
right  triangle  as  homologous  sides,  the  polygon  described  upon 
the  hypothenuse  is  equivalent   to  the  sum  of  the  polygons 
upon  the  other  two  sides. 

ADVANCE  THEOREMS. 

435.  If  two  triangles  have  two  sides  of  one  equal  respec- 
tively to  two  sides  of  the  other,  and  their  included  angles 
supplementary,  the  triangles  are  equivalent. 

436.  If  a  straight  line  be  drawn  through  the  centre  of  a 
parallelogram,  the  two  parts  are  equivalent. 


PLANE  GEOMETRY.  95 

437.  If  through  the  middle  point  of  the  median  of  a  trape- 
zoid a  line  be  drawn,  cutting  the  bases,  the  two  parts  are 
equivalent. 

438.  In  every  trapezoid  the  triangle  which  has  for  its  base 
one  leg,  and  for  its  vertex  the  middle  point  of  the  other  leg,  is 
equivalent  to  one-half  the  trapezoid. 

439.  If  any  point  within  a  parallelogram,  selected  at  ran- 
dom, be  joined  to  the  four  vertices,  the  sum  of  the  areas  of 
either  pair  of  opposite  triangles  is  equivalent  to  one-half  the 
parallelogram. 

440.  The  area  of  a  trapezoid  is  equal  to  the  product  of  one 
of  its  legs  and  the  distance  of  this  leg  from  the  middle  point 
of  the  other. 

441.  The  triangle  whose  vertices  are  the  middle  points  of 
the  sides  of  a  given  triangle  is  equivalent  to  one-fourth  the 
latter. 

442.  The  parallelogram  formed  in  101  is  equivalent  to  one- 
half  the  quadrilateral. 

443.  If  two  parallelograms  have  two  contiguous  sides  respec- 
tively equal,  and  their  included  angles  supplementary,  the  par- 
allelograms are  equivalent. 

444.  The  lines  joining  the  middle  point  of  either  diagonal 
of  a  quadrilateral  to  the  opposite  vertices,  divide  the  quadri- 
lateral into  two  equivalent  parts. 

445.  The  line  which  joins  the  middle  points  of  the  bases  of 
a  trapezoid  divides  the  trapezoid  into  two  equivalent  parts. 

PROBLEMS  OF  COMPUTATION. 

446.  Compute  the  area  of  a  right  isosceles  triangle  if  the 
hypothenuse  is  100  rods. 


96  PLANE  GEOMETRY. 

447.  Compute  the  area  of  a  rhombus  if  the  sum  of  its  diago- 
nals is  12  inches  and  their  ratio  is  3 : 5. 

448.  Compute  the  area  of  a  right  triangle  whose  hypothe- 
nuse  is  13  feet  and  one  of  whose  legs  is  5  feet. 

449.  Compute  the  area  of  an  equilateral  triangle,  one  of 
whose  sides  is  40  feet. 

450.  The  area  of  a  trapezoid  is  3^  acres ;  the  sum  of  the  two 
parallel  sides  is  242  yards.    Find  the  perpendicular  distance 
between  them. 

451.  The  diagonals  of  a  rhombus  are  24  feet  and  40  feet 
respectively.     Compute  its  area. 

452.  The  diagonals  of  a  rhombus  are  88  feet  and  234  feet 
respectively.     Compute  its  area,  and  find  length  of  one  of  its 
sides. 

453.  The  area  of  a  rhombus  is  354,144  square  feet,  and  one 
diagonal  is  672  feet.     Compute  the  other  diagonal  and  one 
side. 

454.  The  sides  of  a  right  triangle  are  in  the  ratio  of  3,  4, 
and  5,  and  the  altitude  upon  the  hypothenuse  is  20  yards. 
Compute  the  area. 

455.  Compute  the  area  of  a  quadrilateral  circumscribed  about 
a  circle  whose  radius  is  25  feet  and  the  perimeter  of  the  quad- 
rilateral 400  feet. 

456.  Compute  the  area  of  a  hexagon  having  the  same  length 
of  perimeter  and  circumscribed  about  the  same  circle. 

457.  The  base  of  a  triangle  is  75  rods  and  its  altitude  60 
rods.     Find  the  perimeter  of  an  equivalent  rhombus  if  its 
altitude  is  45  rods. 


PLANE  GEOMETRY.  97 

458.  Upon  the  diagonal  of  a  rectangle  40  yards  by  25  yards 
an  equivalent  triangle  is  constructed.     Compute  its  altitude. 

459.  Compute  the  side  of  a  square  equivalent  to  a  trapezoid 
whose  bases  are  56  feet  and  44  feet  respectively,  and  each  of 
whose  legs  is  10  feet. 

460.  Find  what  part  of  the  entire  area  of  a  parallelogram 
will  be  the  area  of  the  triangle  formed  by  drawing  a  line  from 
one  vertex  to  the  middle  point  of  one  of  the  opposite  sides. 

461.  In  two  similar  polygons  two  homologous  sides  are  15 
feet  and  25  respectively.     The  area  of  the  first  polygon  is  450 
square  feet.     Compute  the  area  of  the  other  polygon. 

462.  The  base  of  a  triangle  is  32  feet  and  its  altitude  20 
feet.     Compute  the  area  of  the  triangle  formed  by  drawing  a 
line  parallel  to  the  base  at  a  distance  of  15  feet  from  the 
base. 

463.  The  sides  of  two  equilateral  triangles  are  20  and  30 
yards  respectively.     Compute  the  side  of  an  equilateral  tri- 
angle equivalent  to  their  sum. 

464.  If  the  side  of  one  equilateral  triangle  is  equivalent  to 
the  altitude  of  another,  what  is  the  ratio  of  their  areas  ? 

465.  The  radius  of  a  circle  is  15  feet,  and  through  a  point 
9  inches  from  the  centre  any  chord  is  drawn.     What  is  the 
product  of  the  two  segments  of  this  chord  ? 

466.  A  square  field  contains  5f  acres.     Find  the  length  of 
fence  that  incloses  it. 

467.  A  square  field  210  yards  long  has  a  path  round  the 
inside  of  its  perimeter  which  occupies  just  one-seventh  of  the 
whole  field.     Compute  the  width  of  the  path. 

468.  A  street  1J  miles  long  contains  5  acres.     How  wide  is 
the  street  ? 


98  PLANE  GEOMETRY. 

469.  The  perimeter  of  a  rectangle  is  72  feet,  and  its  length 
is  twice  its  breadth.     What  is  its  area  ? 

470.  A  chain  80  feet  long  incloses  a  rectangle  15  feet  wide. 
How  much  more  area  would  it  inclose  if  the  figure  were  a 
square  ? 

471.  The  perimeter  of  a  square,  and  also  of  a  rectangle 
whose  length  is  four  times  its  breadth,  is  400  yards.     Compute 
the  difference  in  their  areas. 

472.  A  rectangle  whose  length  is  25m  is  equivalent  to  a 
square  whose  side  is  15m.     Which  has  the  greater  perimeter, 
and  how  much  ? 

473.  The  perimeters  of  two  rectangular  lots  are  102  yards 
and  108  yards  respectively.     The  first  lot  is  -f  as  wide  as  it  is 
long,  and  the  second  lot  is  twice  as  long  as  it  is  wide.     Com- 
pute the  difference  in  the  value  of  the  two  lots  at  $1  per 
square  foot. 

474.  A  rhombus  and  a  rectangle  have  equal  bases  and  equal 
areas.     Compute  their  perimeters  if  one  side  of  the  rhombus 
is  15  feet  and  the  altitude  of  the  rectangle  is  12  feet. 

475.  The  altitudes  of  two  triangles  are  equal,  and  their 
bases  are  20  feet  and  30  feet  respectively.     Compute  the  base 
of  a  triangle  equivalent  to  their  sum  and  having  an  altitude 
J  as  great. 

REGULAR  POLYGONS  AND  OIEOLES. 
See  328. 

476.  An  equilateral  polygon  inscribed  in  a  circle  is  regular. 
Sug.   See  Theorem  193. 

477.  An  equiangular  polygon  circumscribed  about  a  circle 
is  regular. 


PLANE  GEOMETRY. 


99 


478.  If  a  polygon  be  regular,  a  circle  can  be  circumscribed 
about  it ;  i.e.  one  circumference  can  be  constructed  which  shall 
pass  through  all  its  vertices. 

Post.  Let  ABCDH  be  a  regular 
polygon  of  n  sides. 

We  are  to  prove  that  a  circum- 
ference can  be  constructed  which 
will  pass  through  all  its  vertices. 

Dem.  A  circumference  may  be 
passed  through  any  three  vertices, 
as  A,  B,  and  C.  (See  Theorem 
218.) 

From  the  centre,  K,  of  this 
circumference  draw  lines  to  all  the  vertices. 

Z  ABC  =  Z  BCD.  Why  ? 

What  relation  exists  between  the  two  triangles  AKB  and 
BKC?  Why? 

What  relation,  then,  exists  between  the  two  angles  ABK 
and  BCK?  Why  ? 

What  relation  must  consequently  exist  between  the  two 
angles  KBC  and  KCD  ?  Why  ? 

How,  then,  do  the  two  triangles  KBC  and  KCD  compare  ? 
Why? 

What  must  therefore  be  the  relation  between  KD  and  KC  ? 
Why? 

What  must  be  true,  then,  of  the  circumference  passing 
through  the  vertices  A,  B,  and  C,  as  regards  the  vertex  D  ? 

In  a  similar  manner,  fix  the  position  of  the  circumference 
with  regard  to  each  of  the  other  vertices. 

479.  If  a  polygon  be  regular,  a  circle  may  be  inscribed  in  it ; 
i.e.  a  circle  may  be  constructed  which  shall  have  the  sides  of 
the  polygon  as  tangents. 

Sug.  First  circumscribe  a  circle  by  Theorem  478,  then  con- 
sult 214. 


100  PLANE  GEOMETRY. 

480.   One  side  of  a  regular  hexagon  is  equal  to  the  radius 
of  the  circumscribed  circle. 

Post.  Let  ABCDHKbe  a  regular 
hexagon,  and  let  Nloe  the  centre  of 


the  circumscribed  circle   (Theorem 
£  478),  and  let  NA  and  NKbe  drawn. 

the  are  to  prove  AK=  AN. 

Dem.  The  arc  AK  is  what  part  of 
We  entire  circumference  ?  Why  ? 

What  is  the  value  of  the  angle  N 
then? 

What,  then,  must  be  the  value  of 
H 

/.NAK+Z-NKAt     Why? 

What  relation  exists  between  the  A  NAK  and  NKA  ? 
Why? 

What,  then,  must  be  the  value  of  each  one  of  them  ? 

From  this  relation  of  the  three  angles  what  must  be  the 
relation  of  the  sides  of  the  A  AKN  ? 

Hence  AK  =  AN.  Q.E.D. 

481.  If  the  circumference  of  a  circle  be  divided  into  any 
number  of  equal  arcs,  the  chords  joining  the  successive  points 
of  division  will  form  a  regular  polygon. 

482.  If  the  circumference  of  a  circle  be  divided  into  any 
number  of  equal  arcs,  the  tangents  at  the  points  of  division 
will  form  a  regular  polygon. 

483.  Tangents  to  a  circumference  at  the  vertices  of  a  regular 
inscribed  polygon  form  a  regular  circumscribed  polygon  of  the 
same  number  of  sides  as  the  inscribed  polygon. 

484.  Def.   The  radius  of  the  circumscribed  circle  is  called 
the  radius  of  the  polygon. 

The  radius  of  the  inscribed  circle  is  called  the  apothegm  of  the 


PLANE  GEOMETRY.  101 

polygon.     The  common  centre  of  the   circumscribed  and  in- 
scribed circles  is  also  the  centre  of  the  polygon. 

The  angle  formed  by  two  polygonal  radii  is  called  the 
polygonal  central  angle,  and  that  formed  by  two  sides  simply 
the  polygonal  angles. 

485.  If  all  the  radii  of  a  regular  polygon  of  n  sides  be 
drawn,  how  many  triangles  will  thus  be  formed  ? 

What  kind  of  triangles  will  they  be  ? 

What  will  be  their  relation  to  one  another  as  regards  magni- 
tude? 

Find  an  expression  for  the  central  polygonal  angle  in  terms 
of  n  and  a  right  angle. 

Find  an  expression  for  the  polygonal  angle  in  terms  of  the 
Su-me  quantities.  (See  329.) 

What  relation  exists  between  the  polygonal  angle  and  the 
polgonal  central  angle  ?  Prove. 

=  p0i.  angie,     and     -  =  Cent,  angle. 


n  n 

Sug.   See  above  and  compare  the  two  expressions. 
How  does  the  radius  of  the  polygon  divide  the  polygonal 
angle  ? 

486.  If  radii  be  drawn  to  a  regular  circumscribed  polygon, 
and   chords   of  the  circle  be  drawn   joining  the   successive 
points  of  intersection,  these  chords  will  form  a  regular  poly- 
gon of  the  same  number  of  sides  as  the  circumscribed  polygon, 
and  the  sides  of  the  two  polygons  will  be  parallel,  two  and 
two. 

487.  If  tangents  be  drawn  at  the  middle  points  of  the  sev- 
eral arcs  subtended  by  the  sides  of  a  regular  polygon,  these 
tangents  will  form  a  regular  polygon  whose  vertices  lie  on  the 
radii  extended  of  the  inscribed  polygon,  and  whose  sides  are 
respectively  parallel  to  those  of  the  latter. 


102  PLANE  GEOMETRY. 

488.  If  the  vertices  of  a  regular  inscribed  polygon  of  n 
sides  are  joined  to  the  middle  points  of  the  arcs  subtended  by 
the  sides  of  the  polygon,  the  lines  thus  drawn  will  form  a 
regular  inscribed  polygon  of  2n  sides. 

489.  If  tangents  are  drawn  at  the  middle  points  of  the  arcs 
between  adjacent  points  of  contact  of  the  sides  of  a  regular 
circumscribed  polygon  of  n  sides,  a  regular  circumscribed  poly- 
gon of  2  n  sides  will  thus  be  formed. 

490.  The  perimeter  of  a  regular  inscribed  polygon  of  n 
sides  is  less  than  the  perimeter  of  the  regular  polygon  of  2n 
sides  inscribed  in  the  same  circle. 

491.  The  perimeter  of  the  regular  circumscribed  polygon  of 
n  sides  is  greater  than  the  perimeter  of  the  regular  polygon  of 
2  n  sides  circumscribed  about  the  same  circle. 

492.  Two  regular  polygons  of  the  same  number  of  sides  are 
similar. 

H" 


Post.  Let  P  and  P  be  two  regular  polygons,  each  having  n 
sides.  We  are  to  prove  that  they  are  similar ;  i.e.  that  their 
homologous  sides  form  a  proportion,  and  that  they  are  mutu- 
ally equiangular. 

Dem.  What  is  the  relation  between  all  the  angles  of  poly- 
gon P  ?  Of  polygon  Pf  ?  Why  ? 

I.  What  is  the  value  of  one  of  the  polygonal  angles,  as  -4, 
of  polygon  P  ?  See  329  and  485. 


PLANE  GEOMETRY.  103 

What  is  the  value,  expressed  in  same  terms,  of  one  of  the 
polygonal  angles,  as  JVJ  of  the  polygon  P'  ? 

What,  then,  is  the  relation  between  the  angles  A  and  JV? 

Hence,  what  must  be  true  regarding  the  corresponding 
angles  of  the  two  polygons  ? 

II.    What  is  the  relation  between  AB  and  EG  ?     Why  ? 


Hence 

What  is  the  relation  between  NQ  and  QR  ?     Why  ? 

Therefore  =        .  Why? 


Or  AB  :  BC  :  :  NQ  :  QR,         (proportion  form) 

or  ABiNQiiBC:  QR.  Why  ? 

The  pupil  should  finish  the  demonstration. 

493.  The  areas  of  two  regular  polygons  of  the  same  number 
of  sides  are  in  the  same  ratio   as  the  squares  of  any  two 
homologous  sides. 

Sug.   Consult  431. 

494.  The  perimeters  of  two  regular  polygons  of  the  same 
number  of  sides  are  in  the  same  ratio  as  the  radii  of  their  cir- 
cumscribed circles. 

Sug.   Consult  359  and  492. 

495.  The  perimeters  of  two  regular  polygons  of  the  same 
number  of  sides  are  in  the  same  ratio  as  their  apothegms. 

496.  The  areas  of  two  regular  polygons  of  the  same  number 
of  sides  are  in  the  same  ratio  as  the  squares  of  the  radii  of 
their  circumscribed  circles. 

497.  The  areas  of  two  regular  polygons  of  the  same  num- 
ber of  sides  are  in  the  same  ratio  as  the  squares  of  their 
apothegms. 


104  PLANE  GEOMETRY. 

498.  The  area  of  a  regular  polygon  is  equal  to  one-half  the 
product  of  its  perimeter  and  apothegm. 

Sug.   Draw  radii,  then  consult  485. 

499.  Any  curved  or  polygonal  line  which  envelops  an  arc  of 
a  circle  from  one  extremity  to  another  is  longer  than  the 
enveloped  arc. 


Post.  Let  AMB  be  the  arc  of  a  circle.  Then,  of  all  envelop- 
ing lines,  there  must  be  one  shorter  than  any  of  the  others. 
Left  ACDEB  be  that  line.  (Of  course,  if  there  should  be 
others  equal  in  length  to  ACDEB,  the  argument  would  not 
be  vitiated.  The  essential  condition  is,  that  there  shall  be 
none  shorter.) 

Dem.  There  are  three  possible  relations  between  the  arc 
AMB  and  the  enveloping  line  ACDEB. 

I.  ACDEB  =  AMB. 

II.  ACDEB  <  AMB. 

III.  ACDEB  >  AMB. 

Let  us  suppose  Case  I.  to  be  true.  Then  join  any  two  points 
in  ACDEB  by  a  line  that  shall  not  cut  the  arc  AMB,  as  PQ. 
(The  possibility  of  constructing  this  line  cannot  be  challenged, 
as  ACDEB  envelops  the  arc  AMB.) 

Then,  since  PQ  <  PC  +  CD  +  DQ,  APQED  >  ACDEB. 
But  APQED  is  a  line  that  envelops  the  arc  AMB.  Hence 
we  have  arrived  at  the  result  that  this  enveloping  line  is 
shorter  than  ACDEB.  This  is  in  direct  conflict  with  the 


PLANE  GEOMETRY.  105 

hypothesis,  which  was  that  the  latter  was  the  shortest  en- 
veloping line.  Hence  Case  I.  is  inadmissible.  Case  II.  leads 
to  the  same  result.  Hence  Case  III.  is  alone  true. 

In  the  same  manner  it  can  be  proved  that  any  convex  line 
which  returns  into  itself  is  shorter  than  any  line  enveloping 
it  on  all  sides,  whether  the  enveloping  line  touches  the  given 
convex  line  in  one  or  several  places  or  surrounds  without 
touching  it. 

500.  The  circumference  of  a  circle  is  greater  than  the  perim- 
eter of  any  polygon  inscribed  in  it. 

501.  The  circumference  of  a  circle  is  less  than  the  perimeter 
of  any  polygon  circumscribed  about  it. 

Sug.    Consult  499. 

502.  The  area  of  a  regular  inscribed  polygon  of  2n  sides 
is   greater   than   the   area  of  a  regular  polygon  of  n  sides 
inscribed  in  the  same  circle. 

503.  The  area  of  a  regular  circumscribed  polygon  of  2n 
sides  is  less  than  the  area  of  a  regular  polygon  of  n  sides 
circumscribed  about  the  same  circle. 

504.  If  the  number  of  sides  of  a  regular  inscribed  polygon 
be  continuously  doubled,  its  perimeter  will  as  continuously 
increase   in  length,   and   consequently   approach  nearer  and 
nearer  the  length  of  the  circumference. 

505.  If  the  number  of  sides  of  a  regular  inscribed  polygon 
be  continuously  doubled,  its  apothegm  will  as  continuously 
increase   in  length,   and  consequently   approach  nearer  and 
nearer  the  length  of  the  radius. 

506.  If  the   number  of  sides   of  a  regular  circumscribed 
polygon  be  continuously  doubled,  its  perimeter  will  as  con- 
tinuously decrease  in  length,  and  consequently  approach  nearer 
and  nearer  the  length  of  the  circumference. 


106 


PLANE  GEOMETRY. 


507.  If  the   number  of   sides*  of  a  regular  circumscribed 
polygon  be  continuously  doubled,  its  radius  will  as  continu- 
ously decrease   in  length,  and  consequently  approach  nearer 
and  nearer  the  length  of  the  radius  of  the  circle. 

508.  If  the  number  of  sides  of  a  regular  inscribed  polygon 
be  continuously  doubled,  its  area  will  as  continuously  increase, 
and  consequently  approach  nearer  and  nearer  the  area  of  the 
circle. 

509.  If  the   number   of  sides  of  a   regular   circumscribed 
polygon  be  continuously  doubled,  its  area  will  as  continuously 
decrease,  and  consequently  approach  nearer  and  nearer  the 
area  of  the  circle. 

510.  If,  in  504  to  509  inclusive,  the  number  of  sides  be 
doubled  an/infinite  number  of  times,  the  approach  will  be 
infinitely/near;    iiia.   thay^C^BBafli*&:   hence  a  circle  may 
be  regarded  as  a  regular  polygon  with  an  infinite /number  of 
sides,  with   the   circumference  its   perimeter,  ana  radius  its 
apothegm. 

511.  The  area  of  a  regular  inscribed  polygon  of  2n  sides 
is  a  mean  proportional  between  the  areas  of  two  polygons,  each 
of  n  sides,  one  inscribed  within,  and  the  other  circumscribed 
about,  the  same  circle. 

H  G  I  F  Post.  Let  AB  be  one  side  of  the 
regular  inscribed  polygon  of  n  sides, 
and  EF  one  side  of  the  regular  cir- 
cumscribed polygon  of  n  sides  and 
parallel  to  AB.  Let  C  be  the  centre 
of  the  circle.  Draw  the  radii  CA, 
CG,  and  CB,  G  being  the  point  of 
contact  of  the  tangent  EF.  Then 
CA  and  CB,  if  extended,  will  pass 
through  points  E  and  F  respectively.  At  A  and  B  construct 
the  tangents  AH  and  BI}  and  join  CH.  Then  HI  will  be 


PLANE  GEOMETRY.  107 

one  side  of  the  regular  circumscribed  polygon  of  2n  sides. 
Designate  the  area  of  this  polygon  by  P9. 

Again,  let  p  represent  the  area  of  the  inscribed  polygon 
whose  side  is  AB,  or  the  polygon  of  n  sides ;  P  the  area  of 
the  circumscribed  polygon  of  n  sides,  or  whose  side  is  EF-, 
and  p'  the  area  of  the  inscribed  polygon  of  2n  sides,  and 
whose  side  is  AG. 

We  are  to  prove  p  :  pr : :  p' :  P. 

Dem.   It  is  evident  that  the  areas  of  the  A  ACD,  ACG,  and 

EGG  are  -  -  part  of  the  respective  polygons  p,  p'}  and  P. 
£  71 

Area  A  ACD  :  Area  A  ACG  ::CD:CG', 
and  Area  A  ACD :  Area  A  ACG  : :  p   :  p'. 

.-.  p:p'::CD:CG. 

Again,   Area  A  CAG  :  Area  A  CEG  ::CA:  CE, 
and  Area  A  CAG  :  Area  A  CEG  ::  pr  :  P. 

.\p':P::CA:CE. 

But,  since  AD  is  parallel  to  EG,  the  two  A  CAD  and  CGE 
are  similar. 

Hence  CD:CG::CA:  CE. 

.-.  p:p' :  ip'i  P.  Q.E.D. 

512.  The  area  of  a  regular  circumscribed  polygon  of  2n 
sides  is  equal  to  the  quotient  obtained  by  dividing  twice  the 
product  of  the  areas  of  two  regular  polygons,  each  of  n  sides, 
one  inscribed  within,  and  the  other  circumscribed  about,  the 
same  circle,  by  the  sum  of  the  areas  of  two  regular  polygons, 
one  of  n  and  the  other  of  2n  sides,  both  inscribed  in  the  same 
circle.  (See  diagram  and  postulate  of  previous  theorem.) 

n      -p 

We  are  to  prove  P'  =     .     ,. 
p+p' 


108 


PLANE  GEOMETRY. 


Dem.   Since  CH  bisects  the  /.  EGG, 

OH :HE::CG:  CE.    (See  Theorem  323.) 
But        Area  A  CGH :  Area  A  CHE  ::GH:  HE. 
.-.  Area  A  CGH:  Area  A  CHE  ::CG  :  CE. 
Again,    CG  :  CE  : :  CD  :  CA}     or     CG  :  CE  :  :  CD :  CG. 
But  p:p'::CD:  CG.  (Theorem  511.) 

.-.  p:p'::CG:CE. 

.-.  Area  A  CGH :  Area  A  CHE  ::p:p'. 
Hence 

Area  A  CGH :  Area  A  CGH  +  Area  A  CHE  ::p:p  +p', 
or  Area  A  CGH :  Area  A  CGE  :  :  p  :p+p', 

or  2  Area  A  CGH :  Area  A  CGE  ::2p:p+p', 

or  Area  A  CHI  :  Area  A  CGE  ::2p:p  +p'. 

But  areas  of  A  CHI  and  CGE  are  —  part  respectively  of 
the  polygons  P'  and  P. 


Hence     Area  A  CHI  :  Area  A  CGE  : :  P' 
.'.  P':  P  ::  2p:p+p'. 

o  ^  r> 

Whence  J 


P. 


Q.E.D. 


p+p' 

513.  The  chords  which  join  the  extremities  of  two  per- 
pendicular diameters  form  a 
square. 

Post.  Let  CB,  BD,  DA,  and 
AC  be  chords  of  the  circle 
ACBD  connecting  the  extremi- 
ties of  the  two  perpendicular 
diameters  CD  and  AB. 

We  are  to  prove  that  ACBD 
is  a  square. 

Dem.  What  measures  the 
ZACB? 


PLANE  GEOMETRY. 


109 


What  kind  of  an  angle,  then,  must  it  be  ? 
Determine  the  kind  of  angle  at  B,  D,  and  A. 
Compare  the  lengths  of  AC,  CB,  BD,  and  DA. 
Hence  the  figure  ACBD  is  a  square. 


Q.E.D. 


H 


D 


514.  The  diagonals  of  an  inscribed  square  will  be  diameters 
of  the  circle  and  perpendicular  to  each  other. 

515.  The  tangents  to  a  circle  whose  points  of  contact  are 
the  vertices  of  an  inscribed  square 

will  form  a  square. 

Post.  Let  ABCD  be  a  square 
inscribed  in  the  circle  whose  cen- 
tre is  P;  BD  and  AC  its  diago- 
nals; and  HK,  KN,  NQ,  and  QH, 
tangents  whose  points  of  contact 
are  the  vertices  of  the  inscribed 
square. 

We  are  to  prove  that  HKNQ  is 
a  square. 

Dem.  What  measures  each  of  the  A  K,  N,  Q,  and  H? 

What  kind  of  angles,  then,  must  they  be  ? 

What  is  the  relative  position  of  HK  and  QN?  Of  KN 
and  HK?  (See  Theorem  208.) 

What  is  the  relation  between  .RET  and  KC?  Between  BK 
and  BH? 

What  must  be  the  relation,  then,  between  HK  and  KN  ? 

Hence  HKNQ  is  a  square.  Q.E.D. 

516.  The  area  of  an  inscribed  square  is  equivalent  to  twice 
the  square  of  the  radius.     (Use  diagram  of  previous  theorem.) 

It  is  evident  that  the  area  of  the  right  triangle  DPC  is 
one-half  that  of  the  square  upon  PC,  or  the  radius  of  the 
circle. 

The  area  of  the  square  ABCD  is  equal  to  four  times  the 
area  of  the  triangle  DPC, 


110  PLANE  GEOMETRY. 

or  Area  AZ>PO  =  —  (representing  radius  by  r), 

2 

and  4  x  Area  A  DPC  =2r*=  Area  ABCD.  Q.E.D. 

517.  The  area  of  a  circumscribed  square  is  equal  to  four 
times  the  square  of  the   radius  of  the   circle.     (Use   same 
diagram  as  before.) 

It  can  easily  be  shown  that  the  square  HKNQ  is  composed 
of  four  smaller  squares,  each  of  which  is  the  square  of  the 
radius;  hence 

Area  HKNQ  =  4  r8.  Q.E.D. 

518.  Problem.    From  511  and  512  we  have 

I.  j>'=VpxP    and    II.  P  = 


P+p' 

Letting  p  and  P  represent  inscribed  and  circumscribed 
squares  respectively,  we  have  from  516  and  517, 

p  =  2r*    and    P=4?-2. 
By  substituting  these  values  in  I.,  we  have 

p'=  2.82843  r2. 
Then  by  substituting  values  of  p}  p\  and  P  in  II.,  we  have 

P=  3.31371  r2. 

Thus  we  have  computed  the  areas  of  the  regular  inscribed 
and  circumscribed  octagons  in  terms  of  the  radius  of  the  circle. 

Again,  calling  these  p  and  P,  p'  and  P'  will  be  polygons  of 
sixteen  sides;  and  using  the  formulae  as  before,  the  areas 
of  the  latter  may  be  computed.  Kepeating  the  process,  those 
of  thirty-two,  sixty-four,  etc.,  sides  may  be,  in  like  manner, 
computed. 

Below  are  the  tabulated  results  to  seven  decimal  places  for 
thirteen  doublings  of  the  number  of  sides  of  the  polygons. 


PLANE  GEOMETRY. 


Ill 


No.  Sides. 

Area  of  Inscribed 
Polygon. 

Area  of  Circumscribed 
Polygon. 

4  

2.0000000  r2 

4.0000000^ 

8  

2.8284271  r2 

3.3137085^ 

16  

3.  0614675  r2 

3.1825979^ 

32  

3.1214452  r3 

3.1517240^ 

64  

3.1365485  r2 

3.1441184  r2 

128  

3.1403312^ 

3.1422236  r2 

256  

3.1412773^ 

3.1417504  r2 

512  

3.1415138  r2 

3.1416321  v* 

1024  

3.1415729^ 

3.1416025  r2 

2048  

3.1415877^ 

3.1415951^ 

4096  

3.1415914r* 

3.1415933  r2 

8192  

3.1415923  r2 

3.  1415928  r2 

16384  

3.1415925^ 

3.1415927^ 

32768     

3  1415926  r2 

3.1415926^ 

Hence,  since  the  area  of  the  circle  is  greater  than  that  of 
the  inscribed  polygon  and  less  than  that  of  the  circumscribed 
polygon,  3.1415926  r2  must  be  the  area  of  the  circle  correct 
to  within  less  than  one  tenth-millionth  part  of  r2.  But  by 
continuing  the  process,  the  areas  of  the  two  polygons  may  be 
made  to  agree  to  any  desired  number  of  decimal  places,  and 
therefore  such  result  may  be  taken  as  the  area  of  the  circle 
without  sensible  error.  If  r  be  taken  as  unity,  it  would, 
of  course,  vanish  from  the  expression,  and  consequently 
3.1415926  may  be  taken  as  the  area  of  a  circle  ivhose  radius 
is  unity. 

519.   It  can  easily  be  shown  that  the  two  formulae 


and    II.  P'= 


P+P' 


will  be  true  if  these  quantities  represent  perimeters  instead 


112 


PITANE  GEOMETRY. 


of  areas.  Regarding  them  as  such,  and  using  the  diagram 
in  515,  we  will  compute  the  perimeters  in  terms  of  the 
diameter,  calling  the  latter  D. 

Each  side  of  the  circumscribed  square  is,  of  course,  equal  to 
the  diameter,  and  hence  its  perimeter  is  4  J9. 

In  the  right  triangle  APD, 


—  , 

V2 


— 

V2 


Hence  the  perimeter  of  the  inscribed  square  is  2.8284271  D. 
Using  the  formulae  as  suggested,  and  tabulating  the  results, 
we  have  the  following  : 


No.  Sides. 

Perimeter  of  Inscribed 
Polygon. 

Perimeter  of  Circumscribed 
Polygon. 

4  

2  8284271  D 

4  0000000  D 

8  

3  0614675  D 

3  3137085  D 

16  

3  1214452  D 

3  1825979  D 

32  

3.  1365485  D 

3  1517249  D 

64  

3.  1403312  D 

3  1441184  D 

128  

3  1412773  D 

3  1422236  D 

256  
512  

3.1415138  D 
3  1415729  D 

3.1417504  D 
3  1416321  D 

1024  ....   .  . 

3  1415877  D 

3  1416025  D 

2048  

3.1415914  D 

3  1415951  D 

4096  

3.  1415923  D 

3  1415933  D 

8192  

3.1415924  D 

3  1415928  D 

16384  

3  1415925  D 

3  1415927  D 

32768  

3.1415926  D 

3  1415926  D 

Hence,  since  the  circumference  of  the  circle  is  greater  than 
the  perimeter  of  the  inscribed  polygon,  and  less  than  that  of 


PLANE  GEOMETRY. 


113 


the  circumscribed  polygon,  3. 1415926  D  must  be  the  circum- 
ference of  the  circle  correct  to  within  less  than  one  ten- 
millionth  part  of  D.  But  by  continuing  the  process  the 
perimeters  of  the  two  polygons  may  be  made  to  agree  to 
any  desired  number  of  decimal  places,  and  therefore  such 
result  may  be  taken  as  the  circumference  of  the  circle  without 
sensible  error.  If  D  be  taken  as  unity,  it  would,  of  course, 
vanish  from  the  expression,  and  consequently  3.1415926  may 
be  taken  as  the  circumference  of  a  circle  whose  diameter  is  unity. 

520.   The  circumferences  of  two  circles  are  in  the  same  ratio 
as  their  radii,  and  also  their  diameters. 


Post.  Let  ABHK,  etc.,  and  STVY,  etc.,  be  two  circles 
whose  centres  are  JVand  W,  and  designate  the  circumferences 
by  C  and  c,  and  their  radii  by  R  and  r,  and  their  diameters 
by  D  and  d,  the  capital  letter  referring  to  the  larger  circle. 

We  are  to  prove 

I.  C:c::E:r. 

II.  C:c::D:d. 

Cons.  Inscribe  in  each  a  regular  polygon  of  n  sides,  and 
construct  the  radii  NB  and  WT,  and  the  apothegms  NQ  and 
WE. 

Dem.   Designating  the  perimeters  by  P  and  p, 


P:p::NQ:  WR. 


Why? 


114  PLANE  GEOMETRY. 

If  now  we  inscribe  polygons  with  double  the  number  of 
sides,  and  continue  this  process  indefinitely,  the  perimeters 

will  coincide  with  the  circumferences,  and  the  apothegms  with 
the  radii. 

Hence  I.    C:c::   R  :  r, 

and                                          C:c::2jft:2r;  Why? 

or                                      II.  C :  c  : :   D  :  d.  Q.E.D. 

521.  The  areas  of  two  circles  are  in  the  same  ratio  as  the 
squares  of  their  radii  and  of  their  diameters. 

Sug.   Use  method  similar  to  the  above,  and  consult  496 
and  497. 

521  (a).   Def.  Similar  arcs,  sectors,  and  segments  are  those 
that  correspond  to  equal  central  angles. 

522.  Similar  arcs  are  in  the  same  ratio  as  the  radii  of  the 
circumferences  of  which  they  are  a  part,  and  also  as   the 
diameters. 

Cf 


Post.   Let  CB  and  KH  be  two  similar  arcs,  and  A  and  D 
the  centres  of  the  circumferences  of  which  they  are  a  part. 
Cons.   Draw  the  radii  AC,  AB,  DK,  and  DH. 
(Designate  circumferences,  diameters,  and  radii,  as  before.) 
We  are  to  prove 

I.  ATcCB:ATcKH::K:r. 

II.  ATcCB:AicKH::D:d. 

Dem.  ZA  =  Z.D.  Why? 

Hence  arc  CB  is  the  same  part  of  the  circumference  C,  as 
arc  KH  is  of  the  circumference  c. 

.-.  Arc  CB  :  Arc  KH : :  C  :  c. 


PLANE  GEOMETRY.  115 

But  R:r::C:c.         (See  Theorem  520.) 

.-.I.     ATcCB:A.TcKH::R:r',  Why? 

II.   Arc  05:  AicKH::D:d.  Why? 

Q.E.P. 

523.  The  areas  of  similar  sectors  are  in  the  same  ratio  as 
the   squares  of  the  radii,  and  also  of  the  diameters,  of  the 
circles  of  which  they  are  a  part. 

524.  The  area  of  a  circle  is  equal  to  one-half  the  product  of 
its  circumference  and  radius. 

Sug.   Consult  498  and  510. 

525.  The  area  of  a  sector  is  equal  to  one-half  the  product  of 
its  arc  and  radius. 

526.  The  areas  of  similar  segments  are  in  the  same  ratio 
as  the  squares  of  their  radii,  the  squares  of  their  diameters, 
and  as  the  squares  of  their  chords. 

527.  Let  us  designate  the  circumference  of  a  circle  whose 
diameter  is  unity  by  v,  and  the  circumference  of  any  other 
circle  by  C  ;  its  diameter  by  D  ;  its  radius  by  R  ;  and  its  area 
by  A.  -' 

Then  0:ir::D:l.  Why? 

Hence  I.       O  =  7rZ>; 

C1 

whence  II.     —  =  TT, 

or  III.    C  =  7rx2#. 

TD 

Multiplying  both  members  of  this  equation  by  —  ,  we  have 


CR 
But,  by  524,  ¥—  =  the  area  of  the  circle  ;  hence 

2 

IV.     A  = 


116  PLANE  GEOMETRY. 

528.  Hence  the  area  of  any  circle  is  equal  to  the  square  of  its 
radius  multiplied  by  the  constant  quantity  TT,  and  the  circumfer- 
ence of  every  circle  is  equal  to  the  product  of  its  diameter  (or 
twice  its  radius)  by  the  same  quantity  IT. 

From  II.  above,  it  is  readily  seen  that  TT  is  the  ratio  of  the 
circumference  of  any  circle  to  its  diameter,  or  of  a  circumfer- 
ence to  its  radius. 

The  exact  numerical  value  of  TT  can  be  only  approximately 
expressed.  As  computed  in  352  it  is  3.1415926,  but  for  prac- 
tical purposes  in  computing  its  value  is  usually  taken  as 
3.1416. 

The  symbol  TT  is  the  first  letter  of  the  Greek  word  meaning 
perimeter  or  circumference. 

529.  The  quadrature  of  the  circle  is  the  problem  which  re- 
quires the  finding  of  a  square  which  shall  be  equal  in  area  to 
that  of  a  circle  with  a  given  radius.     Now  since  the  area  of  a 
circle  is  equal  to  its  circumference  multiplied  by  one-half  its 
radius,  if  a  straight  line  of  same  length  as  circumference  be 
taken  as  the  base  of  a  rectangle,  and  one-half  the  radius  as  its 
altitude,  their  product  will  be  the  area  of  the  rectangle,  also 
of  the  circle.     It  is  also  evident  that,  if  a  line  which  is  a  mean 
proportional  between  these  two  be  taken  as  the  side  of  a 
square,  the  area  of  this  square  will  be  equal  to  that  of  the 
rectangle,  and  consequently  to  that  of  the  circle.     It  will  be 
shown  in  the  series  of  problems  of  construction  (Prob.  736) 
how  this  mean  proportional  can  be  found ;  hence,  to  "  square 
the  circle "  we^  must  be  able  to  find  the  circumference  when 
the  radius  is  known,  or  vice  versa.     For  accomplishing  this,  we 
must  know  the  ratio  of  the  circumference  to  its  diameter  or 
radius.     But  this  raticf,  as  has  been  remarked  before,  can  be 
only  approximately  expressed ;  for,  as  the  higher  mathematics 
prove,  the  circumference  and  diameter  are  incommensurable, 
but  the  approximation  has  been  carried  so  far  that  the  error  is 
infinitesimal.     Archimedes,  about  250  B.C.,   was  the   first  to 
assign  an  approximate  value  to  TT.     He  found  that  it  must  be 


PLANE  GEOMETRY.  117 

between  3.1428  and  3.1408.  In  1640  Metius  computed  it  cor- 
rectly to  6  places.  Later,  in  1579,  Vieta  carried  the  approx- 
imation to  10  places,  Van  Ceulen  to  36  places,  Sharp  to  72 
places,  Machin  to  100  places,  De  Lagny  to  128  places,  Ruther- 
ford to  208  places,  and  Dr.  Clausen  to  250  places.  In  1853 
Rutherford  carried  it  to  440  places,  and  in  1873  Shanks  com- 
puted it  to  707  places,  but  these  latter  results  do  not  appear 
to  have  been  verified.  The  following  is  its  value  to  208  places, 
as  computed  by  Rutherford : 

v  =  3.141592653589793238462643383279- 
502884197169399375105820974944- 
592307816406286208998628034825- 
342717067982148086513282306647- 
093844609550582231725359408128- 
484737813920386338302157473996- 
0082593125912940183280651744  +. 

530.  Some  idea  of  the  accuracy  of  the  above  value  may  be 
formed  from  the  following  statement  taken  from  Peacock's 
Calculus  :  "If  the  diameter  of  the  universe  be  100,000,000,000 
times  the  distance  of  the  sun  from  the  earth  (about  93,000,000 
miles),  and  if  a  distance  which  is  100,000,000,000  times  this 
diameter  be  divided  into  parts,  each  of  which  is  one  100,000,- 
000,000th  part  of  an  inch ;  then  if  a  circle  be  described  whose 
diameter  is  100,000,000,000  times  that  distance,  repeated  100,- 
000,000,000  times  as  often  as  each  of  those  parts  of  an  inch 
is  contained  in  it ;  then  the  error  in  the  circumference  of  this 
circle,  as  computed  from  this  approximation,  will  be  less  than 
one  100,000,000,000th  part  of  the  one  100,000,000,000th  part 
of  an  inch." 

ADVANCE  THEOREMS. 

531.  In  two  circles  of  different  radii,  angles  at  the  centres 
subtending  arcs  of  equal  length  are  to  each  other  inversely  as 
the  radii. 


118  PLANE  GEOMETRY. 

532.  If,  from  any  point  within  a  regular  polygon  of  n  sides, 
perpendiculars  be  drawn  to  all  the  sides,  the  sum  of  these  per- 
pendiculars is  equal  to  n  times  the  apothegm. 

533.  If  perpendiculars  be  drawn  from  the  vertices  of  a  regu- 
lar polygon  to  any  diameter  of  the  circumscribed  circle,  the 
sum  of  the  perpendiculars  upon  one  side  of  the  diameter  is 
equal  to  the  sum  of  those  on  the  other  side. 

534.  An  equiangular  polygon  inscribed  in  a  circle  is  regular 
if  the  number  of  its  sides  be  odd. 

535.  An  equilateral  polygon  circumscribed  about  a  circle  is 
regular  if  the  number  of  its  sides  be  odd. 

536.  The  sum  of  the  squares  of  the  lines  joining  any  point 
in  the  circumference  of  a  circle  with  the  vertices  of  an  inscribed 
square  is  equal  to  twice  the  square  of  the  diameter  of  the 
circle. 

537.  The  area  of  the  ring  included  between  two  concentric 
circles  is  equal  to  the  area  of  the  circle  whose  diameter  is  that 
chord  of  the  outer  circle  which  is  tangent  to  the  inner. 

538.  If  the  sides  of  a  right  triangle  be  the  homologous  sides 
of  similar  polygons,  the  area  of  the  polygon  on  the  hypothe- 
nuse  is  equal  to  the  sum  of  the  areas  of  the  other  two. 

539.  If  three  circles  be  described  upon  the  sides  of  a  right 
triangle  as  diameters,  the  area  of  that  described  upon  the 
hypothenuse  is  equal  to  the  sum  of  the  areas  of  the  other  two. 

540.  If  upon  the  legs  of  a  right  triangle  semi-circumferences 
are  described  outwardly,  the  sum  of  the  areas  contained  be- 
tween these  semi-circumferences  and  the  semi-circumference 
passing  through  the  three  vertices  is  equal  to  the  area  of  the 
triangle. 

541.  If  the  diameter  of  a  circle  be  divided  into  two  parts, 
and  upon  these  parts  semi-circumferences  are   described  on 


PLANE  GEOMETRY.  119 

opposite  sides  of  the  diameter,  these  semi-circumferences  will 
divide  the  circle  into  two  parts  which  have  the  same  ratio  as 
the  two  parts  of  the  diameter. 

542.  If  two  chords  of  a  circle  be  perpendicular  to  each  other, 
the  sum  of  the  areas  of  the  four  circles  described  upon  the  four 
segments  as  diameters  will  be  equal  to  the  area  of  the  given 
circle. 

543.  If  squares  be  constructed  outwardly  upon  the  sides  of 
a  regular  hexagon,  their  exterior  vertices  will  be  the  vertices 
of  a  regular  dodecagon. 

544.  The  radius  of  a  regular  inscribed  polygon  is  a  mean 
proportional  between  its  apothegm  and  the  radius  of  a  regular 
polygon  of  double  the  number  of  sides  circumscribed  about 
the  same  circle. 

545.  The  area  of  a  regular  dodecagon  is  equal  to  three  times 
the  square  of  its  radius. 

546.  If  the  radius  of  a  circle  be  divided  in  extreme  and 
mean  ratio,  the  larger  part  will  be  the  side  of  a  regular  decagon 
inscribed  in  the  same  circle. 

547.  The  apothegm  of  a  regular  inscribed  pentagon  is  equal 
to  one-half  the  sum  of  the  radius  of  the  circle  and  the  side  of 
the  regular  decagon  inscribed  in  the  same  circle. 

548.  The  square  of  the  side  of  a  regular  inscribed  pentagon 
is  equivalent  to  the  sum  of  the  squares  of  the  side  of  the  regu- 
lar inscribed  decagon  and  the  radius  of  the  circle. 

PROBLEMS  OF  COMPUTATION. 

549.  Compute  the  side  of  a  regular  inscribed  trigon  in  terms 
of  the  regular  trigon  circumscribed  about  the   same  circle. 
Compare  their  areas. 

550.  Compute  the  side  of  an  inscribed  square  in  terms  of 
the  square  circumscribed  about  the  same  circle. 


120  PLANE  GEOMETRY. 

551.  Compute  the  apothegm  of  a  regular  inscribed  trigon  in 
terms  of  the  regular  hexagon  inscribed  in  the  same  circle. 

552.  Compute  the  apothegm  of  a  regular  inscribed  hexagon  in 
terms  of  the  regular  trigon  inscribed  in  the  same  circle. 

553.  Regular  trigons  and  hexagons  are  both  inscribed  and 
circumscribed  about  the  same  circle.     Compare  their  areas. 

554.  Compute   the  area  of  a  regular  polygon  of  24  sides 
inscribed  in  a  circle  whose  radius  is  10  inches. 

555.  Compute  the  perimeter  of  a  regular  pentagon  inscribed 
in  a  circle  whose  radius  is  12  feet. 

556.  The  perimeter  of  a  regular  hexagon  is  480  feet,  and 
that  of  a  regular  octagon  is  the  same.     Which  has  the  greater 
area,  and  how  much  ? 

557.  If  paving  blocks  are  in  the  shape  of  regular  polygons 
(i.e.  their  cross-sections),  how  many  shapes  can  be  employed 
in  order  to  completely  fill  the  space  ? 

558.  Compute  the  diameter  of  a  circle  whose  circumference 
is  12  feet  and  10  inches. 

559.  The  diameter  of  a  carriage  wheel  is  4  feet  and  3  inches. 
How  many  revolutions  does  it  make  in  traversing  one-fourth 
of  a  mile  ? 

560.  What  is  the  width  orf  the  ring  between  two  concentric 
circumferences  whose  lengths  are  480  feet  and  360  feet  ? 

561.  Find  the  length  of  an  arc  of  36°  in  a  circle  whose  diam- 
eter is  36  inches. 

562.  In  raising  water  from  the  bottom  of  a  well  by  means 
of  a  wheel  and  axle,  it  was  found  that  the  axle,  whose  diameter 
was  8  inches,  made  20  revolutions  in  raising  the  bucket.     Com- 
pute the  depth  of  the  well. 

563.  Find  the  central  angle  subtending  an  arc  6  feet  and  4 
inches  long,  if  the  radius  of  the  circle  be  8  feet  and  2  inches. 


PLANE  GEOMETRY.  121 

564.  If  the  radius  of  a  circle  is  5  feet  and  3  inches,  find  the 
perimeter  of  a  sector  whose  angle  is  45°. 

565.  If  the  central  angle  subtending  an  arc  10  feet  and  6 
inches  long  is  72°,  what  is  the  length  of  the  radius  of  the  circle  ? 

566.  If  the  length  of  a  meridian  of  the  earth  be  40,000,000 
metres,  what  is  the  length  of  an  arc  of  1"  ? 

567.  Two   arcs  have  the   same   angular  measure,  but  the 
length  of  one  is  twice  that  of  the  other.     Compare  the  radii 
of  those  arcs. 

568.  Compute  the  area  of  a  circle  whose  circumference  is 
100  yards. 

569.  Two  arcs   have   the   same   length,  but  their  angular 
measurements  are  20°  and  30°  respectively.     If  the  radius  of 
the  first  arc  is  6  feet,  compute  the  radius  of  the  other. 

570.  Find  the  circumference  of  a  circle  whose  area  is  2 
acres  and  176  square  yards. 

571.  The  diameter  of  a  circle  is  40  feet.     Find  the  side  of  a 
square  which  is  double  the  area  of  the  circle. 

572.  The  area  of  a  square  is  196  square  rods.     Find  the  area 
of  the  circle  inscribed  in  the  square. 

573.  A  circular  fish-pond  which  covers  an  area  of  5  acres 
and  100  square  rods  is  surrounded  by  a  walk  5  yards  wide. 
Compute  the  cost  of  gravelling  the  walk  at  6J  cents  per  square 
yard. 

574.  What  must  be  the  width  of  a  walk  around  a  circular 
garden  containing  If  acres,  in  order  that  the  walk  may  contain 
exactly  one-fourth  of  an  acre  ? 

575.  A  carpenter  has  a  rectangular  piece  of  board  15  inches 
wide  and  20  inches  long,  from  which  he  wishes  to  cut  the 
largest  possible  circle.     How  many  square  inches  of  the  board 
must  he  cut  away  ? 


122  PLANE  GEOMETRY. 

576.  The  perimeters  of  a  circle,  a  square,  and  a  regular 
trigon,  are  each  equal  to  144  feet.     Compare  their  areas. 

577.  If  the  radius  of  a  circle  be  12  inches,  what  is  the  radius 
of  a  circle  10  times  as  large  ? 

578.  What  will  it  cost  to  pave  a  circular  court  30  feet  in 
diameter,  at  54  cents  per  square  foot,  leaving  in  the  centre  a 
hexagonal  space,  each  side  of  which  measures  4  feet  ? 

579.  A  circle  18  feet  in  diameter  is  divided  into  three  equiv- 
alent parts  by  two  concentric  circumferences.     Find  the  radii 
of  these  circumferences. 

580.  If  the  chord  of  an  arc  be  720  feet,  and  the  chord  of  its 
half  be  369  feet,  compute  the  diameter  of  the  circle. 

581.  The  chord  of  half  an  arc  is  17  feet,  and  the  height  of 
the  arc  7  feet.     Compute  the  diameter  of  the  circle. 

582.  The  radius  of  a  circle  is  12  feet ;  the  chords  which  sub- 
tend two  contiguous  arcs  are  6  feet  and  9  feet  respectively. 
Compute  the  chord  subtending  the  arc  equal  to  the  sum  of  the 
other  two. 

583.  The  lengths  of  two  chords,  drawn  from  the  same  point 
in  the  circumference  of  a  circle  to  the  extremities  of  a  diam- 
eter, are  6  feet  and  8  feet  respectively.     Compute  the  area  of 
the  circle. 

584.  The  chord  of  an  arc  is  32  inches,  and  the  radius  of  the 
circle  is  34  inches.  •  Compute  the  length  of  the  arc. 

585.  The  diameter  of  a  circle  is  106  feet.     Compute  the 
lengths  of  the  two  arcs  into  which  a  chord  90  feet  long  divides 
the  circumference. 

586.  The  area  of  a  sector  is  385  square  feet,  and  the  angle 
of  the  sector  is  36°.     Compute  the  radius  of  the  circle  and 
perimeter  of  the  sector. 


PLANE  GEOMETRY.  123 

587.  Compute  the  area  of  a  segment,  if  the  chord  of  the  arc 
is  56  feet  and  the  radius  of  the  circle  is  35  feet, 

588.  A  room  20  feet  long  and  15  feet  wide  has  a  recess  at 
one  end  in  the  shape  of  the  segment  of  a  circle,  the  chord  being 
15  feet,  and  its  greatest  width  4  feet.     Compute  the  area  of 
the  entire  room. 

589.  Compute  the  number  of  square  feet  of  brick  that  would 
be  required  in  blocking  up  one  of  the  arches  of  a  railway  via- 
duct, if  the  span  of  the  arch  is  60  feet,  height  above  the  piers 
20  feet,  and  distance  from  the  ground  to  the  spring  of  the 
arch  20  feet. 

590.  Compute  the  area  of  a  circle  in  which  the  chord,  3  feet 
long,  subtends  an  arc  of  120°. 

591.  Compute  the  area  of  a  segment  whose  arc  is  300°,  the 
radius  of  the  circle  being  20  inches. 

592.  The  areas  of  two  concentric  circles  are  as  5  to  8.     The 
area  of  that  part  of  the  ring  which  is  contained  between  two 
radii  making  the  angle  45°  is  300  square  feet.     Compute  the 
radii  of  the  two  circles. 

593.  What  is  the  altitude  of  a  rectangle  equivalent  to  a 
sector  whose  radius  is  15  feet,  if  the  base  of  the  rectangle  is 
equal  to  the  arc  of  the  sector  ? 

594.  Compute  the  radius  of  a  circle,  if  its  area  is  doubled  by 
increasing  its  radius  one  foot. 

595.  The  radius  of  a  circle  is  10  inches.     Through  a  point 
exterior  to  the  circle  two  tangents  are  drawn,  making  an  angle 
of  60°.     Compute  the  area  of  the  figure  bounded  by  the  tan- 
gents and  the  intercepted  arc. 

596.  Three  equal  circles  are  drawn  tangent  to  each  other, 
with  a  radius  of  12  feet.     Compute  the  area  contained  between 
the  circles. 


124  PLANE  GEOMETRY. 

597.  Upon  each  side  of  a  square,  as  a  diameter,  a  semi- 
circumference  is  described  within  the  square.     If  the  side  of 
the  square  is  10  inches,  find  the  sum  of  the  areas  of  the  four 
leaves. 

598.  In  a  circle  whose  radius  is  100  feet  two  parallel  chords 
are  drawn  on  the  same  side  of  the  centre,  one  equal  to  the  side 
of  a  regular  hexagon,  and  the  other  to  the  side  of  a  regular 
trigon,  both  inscribed  in  the  given  circle.     Compute  the  area 
of  the  circle  comprised  between  the  two  parallel  chords. 


MAXIMA  AND  MINIMA, 

599.  Def.   Among  quantities  of  the  same  kind,  that  which 
is    greatest   is    called   the   maximum  (plural,   maxima),   and 
that  which  is  the  smallest  is   called  the  minimum    (plural, 
minima). 

For  example :  of  all  lines  inscribed  in  the  same  circle,  the 
diameter  is  the  greatest,  and  therefore  is  a  maximum;  and  of 
all  lines  drawn  from  the  same  point  to  a  given  straight  line, 
that  which  is  perpendicular  is  the  shortest,  and  is  therefore  a 
minimum. 

Def.  If  two  or  more  plane  figures  have  equal  perimeters, 
they  are  said  to  be  isoperimetric. 

Def.  A  plane  figure  is  said  to  be  a  maximum  or  a  minimum 
when  its  area  is  a  maximum  or  a  minimum. 

600.  If  any  number  of  triangles  have  the  same  or  equal 

bases  and  equal  areas,  that  which  is 
isosceles  has  the  minimum  perim- 
eter. 

Post.  Let  ABC  and  ADB  be  two 
triangles  having  the  same  base 
AB  and  equal  areas,  and  let  ACB 
be  isosceles,  having  CA  equal  to 
CB. 


PLANE  GEOMETRY.  125 

We  are  to  prove 

AC  +  CB  +  AB  <  AD  +  DB  +  AB. 
Or,  since  AB  =  AB, 

we  are  to  prove 

AC  +  CB  <AD  +  DB. 

Cons.  From  B  construct  a  perpendicular  to  AB,  and  extend 
it  to  meet  AC  extended  in  K.  Join  DK  and  draw  CH 
through  point  D. 

Dem.  How  must  the  altitudes  of  the  two  A  ACB  and  ADB 
compare  ? 

What  must  be  the  position,  then,  of  the  line  CH  relative  to 
AB? 

How,  then,  do  the  two  A  HCB  and  CBA  compare  ?  Why  ? 
The  two  A  KCH  and  CAB  ?  Why  ? 

How,  then,  must  the  two  A  KCH  and  HCB  compare  ? 
Why? 

What  is  the  position  of  CH  relative  to  KB  ?     Why  ? 

How,  then,  must  OfiT  and  CB  compare  ?     DK  and  DB  ? 

Now  compare  AC  +  CK  with  AD  4-  DK,  and  the  pupil 
should  be  able  to  write,  or  give  orally,  a  complete  and  accurate 
demonstration  of  this  theorem. 

601.  If  any  number  of  triangles  have  the  same  area,  that 
which  is  equilateral  has  the  minimum  perimeter. 

602.  If  any  number  of  triangles  have  the  same  or  equal 
bases   and  equal  perimeters,  that  which  is   isosceles   is  the 
maximum. 

/Sug.  Through  the  vertex  of  the  isosceles  triangle  draw  a 
line  parallel  to  the  base.  Then  prove  that  the  vertex  of  the 
other  triangle  cannot  fall  on  that  line.  Then  compare  their 
altitudes,  and  consequently  their  areas. 

603.  If  any  number  of  triangles  be  isoperimetric,  that  which 
is  equilateral  is  the  maximum. 


126 


PLANE  GEOMETRY. 


604.  If  any  number  of  triangles  have  two  sides  in  each 

respectively  equal,  that  in 
which  these  sides  are  per- 
pendicular to  each  other  is 
the  maximum. 

Sug.  Place  them  so  that 
one  set  of  equal  sides  shall 
coincide,  as  AB.  Then  com- 
pare their  altitudes  DK,  CA,  and  HN,  and  consequently  their 
areas. 

605.  If  any  number  of  equivalent  parallelograms  have  the 
same  or  equal  bases,  the  perimeter  of  that  which  is  rectangu- 
lar is  the  minimum. 

606.  Of  all  rectangles  of  given  area,  the  perimeter  of  the 
square  is  a  minimum. 

607.  If  any  number  of  triangles  have  the  same  or  equal 
bases  and  the  same  or  equal  altitudes,  the  perimeter  of  that 
which  is  isoceles  is  a  minimum. 

608.  Every  closed  plane  figure  of  given  perimeter  whose 

area  is  a  maximum,  must  be  convex. 

Post.   Let  ACBN  be  a  plane  concave 
figure,  with  the  straight  line  AB  which 
''  joins  two  of  its  points  in  its  perimeter 
lying  without. 

We  are  to  prove  that  ACBN  cannot  be 
a  maximum. 

Dem.  Conceive  the  figure  CAB  to  be  revolved  about  AB  as 
an  axis  till  it  comes  to  the  position  AC'B.  Then  the  figures 
ACBN  and  ^IC'jBJVhave  equal  perimeters,  but  the  area  of  the 
latter  will  exceed  that  of  the  former.  Hence  ACBN  cannot 
be  a  maximum  among  isoperimetrical  figures.  But  ACBN  is 
any  concave,  i.e.  non-convex,  plane  figure.  Therefore,  of  all 
isoperimetrical  plane  figures,  the  maximum  must  be  convex. 


PLANE  GEOMETRY. 


127 


E1 


609.  Of  all  plane  figures. that  are  isoperimetric,  that  which 
is  a  circle  is  the  maximum. 

Dem.  I.  It  is  evident  that, 
with  a  given  perimeter,  an  in- 
definite number  of  figures  of 
different  shapes  and  areas  may 
be  constructed.  It  is  also  evi- 
dent that  we  can  diminish  the 
area  indefinitely,  but  cannot 
thus  increase  it.  Consequently, 
there  must  be  among  all  these 
figures  having  the  same  perim- 
eter either  one  maximum  figure, 
or  several  maximum  figures  of  different  forms  and  equal 
areas. 

II.  Let  ACBFG  be  a  maximum  figure  with  a  given  perim- 
eter ;  then  by  608  it  must  be  convex.  Let  also  the  line  AB 
divide  the  perimeter  into  halves ;  then  it  must  also  divide  the 
area  into  halves.  For  suppose  one  of  the  parts,  as  AFB,  to  be 
greater  than  the  other,  and  conceive  this  part  to  be  revolved 
on  AB  as  an  axis  until  it  comes  into  the  same  plane  with 
ACB,  and  let  AF'B  be  its  position  after  revolution.  Hence 
the  perimeter  of  the  figure  AF'BEGA  is  equal  to  that  of  the 
figure  ACBFG  A,  but  the  area  of  the  former  is  greater  than 
that  of  the  latter.  Therefore  the  figure  ACBFG  cannot  be  a 
maximum.  But  by  hypothesis  it  is  a  maximum.  Hence  AB 
must  bisect  the  area  of  ACBFG. 

Since  ACBFG  is  a  maximum,  and  AB  bisects  the  area,  it 
follows  that  the  figure  AF'BFG  is  also  a  maximum. 

Again,  let  F  be  any  point  in  BEG  A  selected  at  random,  and 
F1  its  position  after  revolution.  Join  FF1,  FB,  FA,  F'B,  and 
F'A.  Then  AF=AF',  and  FQ  =  F*Q.  Hence  the  two  tri- 
•angles  AQF  and  AQF'  are  equal. 

Therefore  FF'  is  perpendicular  to  AB. 

Similarly  the  two  triangles  AF'B  and  AFB  are  equal. 


128  PLANE  GEOMETRY. 

The  triangle  AFB  must  be  a  maximum,  otherwise  its  area 
could  be  increased  without  increasing  its  perimeter ;  i.e.  with- 
out increasing  the  lengths  of  the  two  chords  .AF  and  FB,  which 
would  consequently  leave  the  areas  of  the  two  segments  AGF 
and  FEE  unchanged,  and  therefore  make  up  an  area  greater 
than  ABEG,  by  which  it  is  evident  that  ACBFG  could  not 
be  a  maximum;  but  this  also  conflicts  with  the  hypothesis 
which  grants  that  ACBFG  is  a  maximum.  Consequently  the 
triangle  AFB  must  be  a  maximum,  and  therefore  the  angle 
AFB  must  be  a  right  angle.  (See  Theorem  604.)  But  F  is 
any  point  in  the  curve  BEFGA.  Hence  BFA  must  be  a  semi- 
circle, as  also  ACB.  Hence  the  whole  figure  ACBFG  must 
be  a  circle.  Q.E.D. 

610.  Of  all  plane  figures  having  equal  areas,  the  perimeter 
of  that  which  is  a  circle  is  the  minimum. 


Post.  Let  C  be  a  circle,  and  A  any  other  plane  figure  having 
the  same  area. 

We  are  to  prove  Perimeter  C  <  Perimeter  A. 

Dem.  For  let  B  be  a  circle  having  a  perimeter  equal  to  that 
of  A.  Hence  by  609  area  B  is  greater  than  area  A,  and  hence 
greater  than  area  C.  Hence  if  area  C  is  less  than  area  B, 
what  must  be  true  of  their  perimeters,  i.e.  their  circumfer- 
ences ?  But  by  construction 

Perimeter  B  =  Perimeter  A. 
Hence  perimeter  of  C  is  a  minimum.  Q.E.D. 


PLANE  GEOMETRY. 


129 


611.   Of  all  mutually  equilateral  polygons,  that  which,  can 
be  inscribed  in  a  circle  is  the  maximum. 


Post.  Let  ABCDH  =  P  and  A'B'C'D'H'  =  P'  be  two  mutu- 
ally equilateral  polygons,  having  AB  equal  to  A'B',  BC  equal 
to  B'C',  etc.,  of  which  ABCDH  can  be  inscribed  in  a  circle, 
and  let  N  be  the  centre  of  such  circle. 

Cons.  With  the  radius  NB  construct  the  arcs  A'B1,  B'C', 
C'D',  etc. 

Dem.   The  Arc  AB  =  Arc  A'B',  and  Arc  BC=  Arc  B'C',  etc. 

Hence  Circumference  ABCDH =  sum  of  the  Arcs  A'B',  B'C1, 
etc. 

Hence        Perimeter  of  S  =  Perimeter  of  S1. 

Therefore  Area  S  >  Area  S'.  (Theorem  609.) 

But  the  corresponding  segments  are  equal.  Hence,  sub- 
tracting their  respective  sums  from  the  above  inequality  leaves 

Area  P  >  Area  P1.  Q.E.D. 

612.  Of  all  isoperimetric  poly- 
gons of  the  same  number  of 
sides,  that  which  is  equilateral 
is  the  maximum. 

Post.  Let  ABCDHK  be  the 
maximum  of  all  isoperimetrical 
polygons  of  any  given  number 
of  sides. 


130  PLANE  GEOMETRY. 

We  are  to  prove  that  it  is  equilateral ;  i.e.  that 
KH=HD=DC,   etc. 

Cons.   Connect  any  two  alternate  vertices,  as  AH. 

Dem.  The  AAKH  must  be  a  maximum  of  all  isoperimet- 
rical  triangles  having  the  common  base  AH\  otherwise  another 
triangle,  as  ANH,  could  be  constructed,  having  the  same  perim- 
eter and  a  greater  area,  in  which  case  the  area  of  the  polygon 
ABCDHN  would  be  greater  than  that  of  ABCDHK.  Hence 
the  latter  would  not  be  a  maximum.  This  result  conflicts  with 
our  hypothesis,  which  grants  that  it  is  a  maximum.  Therefore 
the  triangle  AKH  must  be  a  maximum,  and  consequently 
isosceles.  (See  Theorem  602.) 

Hence  AK=  KH. 

Similarly,  by  joining  KD, 

KH=  HD,  etc. 

Hence  the  polygon  is  equilateral.  Q.E.D. 

613.  The  maximum  of  all  equilateral  polygons  of  the  same 
number  of  sides  is  that  which  is  regular. 

614.  Of  all  polygons  having  the  same  number  of  sides  and 

equal  areas,  the  perimeter  of  that  one 
which  is  regular  is  a  minimum. 

Post.  Let  P  be  a  regular  polygon, 
and  M  any  irregular  polygon  having 
the  same  number  of  sides  and  same 
area  as  P;  and  let  N  be  a  regular 
polygon  having  the  same  number  of 
sides  and  isoperimetrical  with  M. 


PLANE  GEOMETRY. 


131 


(See  612  and  613.) 
Why? 


Dem.  Area  M<N. 

Area  M  —  Area  P. 
.-.  Area  P  <  N. 

But  of  two  regular  polygons  of  the  same  number  of  sides, 
that  which  has  the  less  area  must  have  the  less  perimeter. 
Why? 

Hence  Perimeter  P<  Perimeter  N, 

and  .  •.  Perimeter  P  <  Perimeter  M. 

Hence  perimeter  of  P  is  a  minimum.  Q.E.D. 

615.  Of  all  isoperimetric  regular  polygons,  that  which  has 
the  greatest  number  of  sides  is  the  maximum. 


-A  J)  B 

Post.   Let  ABC  be  a  regular  trigon,  and  P  a  regular  tetragon, 
having  equal  perimeters. 

We  are  to  prove    Area  P  >  Area  ABC. 

-   Cons.   Draw  from    C  any  line    CD  to   AB.  At    C  make 

Z  DCH  equal  to   the  Z  CD.B,   and    <7ZT  equal  to   DB,  and 
join  .HZ). 

Dem.                        A  <mff  =  A  CDB.  Why  ? 

Hence  Area  ABC  =  Area  AMTC.  Why  ? 

But  Area  P  >  Area  ADHC.  Why  ? 

Hence  Area  P  >  Area  AB(7. 

Similarly,  P  could  be  shown  to  be  less  than  an  isoperimetric 
regular  pentagon,  etc.  Q.E.D. 


132  PLANE  GEOMETRY. 

616.  The  area  of  a  circle  is  greater  than  the  area  of  any 
polygon  of  equal  perimeter. 

617.  Of  all  regular  polygons  having  a  given  area,  the  perim- 
eter of  that  which   has  the  greatest  number   of  sides   is  a 
minimum. 


Post.  Let  Q  and  P  be  two  regular  polygons  having  equal 
areas,  and  Q  having  the  greater  number  of  sides. 

We  are  to  prove  that  the  perimeter  of  P  is  greater  than 
that  of  Q. 

Dem.  Let  R  be  a  regular  polygon  whose  perimeter  is  equal 
to  that  of  Q,  but  the  number  of  sides  the  same  as  P. 

Then  Q  >  R.  Why  ? 

But  Area  Q  =  Area  P. 

.:  Area  P  >  Area  R. 

.•.  Perimeter  P>  Perimeter  R.  Why? 

But  Perimeter  R  =  Perimeter  Q. 

.-.  Perimeter  P  >  Perimeter  Q. 
Hence  perimeter  Q  is  a  minimum.  Q.E.D. 

618.  The  circumference  of  a  circle  is  less  than  the  perimeter 
of  any  polygon  of  equal  area. 

619.  The  rectangle  formed  by  the  two  segments  of  a  line 
is  maximum  when  the  segments  are  equal. 


PLANE  GEOMETRY. 

PROBLEMS  OF  CONSTRUCTION. 

620.  In  the  demonstration  of  the  foregoing  theorems  it  has 
been  assumed  that  certain  constructions  were  possible;   i.e. 
that  perpendiculars  and  parallels  could  be  drawn,  that  lines 
and  angles  could  be  bisected,  etc.     It  is  now  proposed   to 
show  that  those  and  many  other  problems  can  be  performed, 
so  our  previous  demonstrations  are  not  vitiated  that  are  in 
any  way  dependent  upon  such  constructions. 

621.  The  solution  of  a  geometrical  problem  of  construction 
involves  in  general  three  steps;  viz.: 

I.  The  construction  proper,  by  the  use  of  the  compass  and 
ruler. 

II.  Demonstration  to  prove  the  correctness  of  the  construc- 
tion. 

III.  Discussion  of  its  limitations  and  applications,  including 
the  number  of  possible  constructions. 

If  numerical  or  algebraical  results  are  also  required,  then 
there  is,  of  necessity, 

IV.  Computation,  by  which  numerical  values  are  ascertained, 
involving  also  the  use  of  general  symbols  in  obtaining  algebraic 
formulae. 

622.  Each  pupil  should  be  provided  with  a  good  pair  of 
compasses,  for  the  use  of  either  ink  or  pencil,  a  good  ruler 
with  straight  edges,  besides  one  hard  and  one  soft  pencil.     The 
lead  of  the  pencil  should  be  sharpened  flat,  so  that  a  fine  line 
can  be  made. 

PLANE  PKOBLEMS. 

623.  It  is  required  to  find  a  point  which  is  a  given  distance 
from  a  given  point. 

Post.   Let  C  be  the  given  point,  and  AB  the  given  dis- 
tance. 


134  PLANE  GEOMETEY. 

We  are  required  to  find  a  point  which  shall  be  at  the  dis- 
tance AB  from  C. 


A  B 


Cons.  First,  with  the  ruler,  from  C  draw  an  indefinite 
straight  line,  as  CD,  in  any  direction.  Then  with  the  com- 
passes, using  C  as  a  centre  and  AB  as  a  radius,  draw  an  arc 
cutting  the  line  CD,  as  at  H. 

Then  H  is  the  required  point ;  for, 

Dem.  If,  in  applying  the  compasses  to  AB  a  circle  had 
been  constructed,  and  the  circle  HKN  also  completed,  then 
the  circles  would  have  equal  radii. 

.-.  His  the  same  distance  from  C  that  A  is  from  B. 

Discussion.  Since,  from  the  definition  of  a  circle,  all  points 
in  the  circumference  are  equally  distant  from  the  centre,  it 
follows  that  every  point  in  the  circumference  IIKN  is  the 
same  distance  from  C  that  A  is  from  B.  Consequently  any 
point  in  that  circumference  answers  the  conditions  of  the 
problem.  Q.E.F. 

623  (a).  Whenever  a  line  is  found  such  that  any  point  in  it 
selected  at  random  will  fulfil  certain  specified  conditions,  or 
such  that  all  points  in  it  have  a  common  property,  that  line 
is  called  the  LOCUS  of  that  point  or  points.  Hence  we  may  say, 
in  the  above  case,  that  the  circumference  HKN  is  the  locus 
of  the  point  which  is  the  distance  AB  from  point  (7,  or,  as 
some  prefer  to  put  it,  it  is  the  locus  of  all  points  which  are 
at  the  distance  AB  from  point  C. 


PLANE  GEOMETRY. 


135 


624.   It   is   required  to  find  the  point,   having   given   its 
distances  from  two  given  points. 


A    B 


Post.  Let  the  two  given  distances  be  A  and  B,  and  the  two 
given  points  C  and  D. 

We  are  required  to  find  a  point  which  is  A  distant  from  C, 
and  B  distant  from  D,  or  vice  versa. 

Sug.  Find  locus  of  the  point  which  is  A  distant  from  point 
(7,  and  locus  of  the  point  which  is  B  distant  from  point  D,  and 
vice  versa. 

Dis.  How  many  points,  then,  satisfy  the  condition  of  the 
problem  ? 

Determine  the  result  if 

I.  The  distance  between  C  and  D  had  been  greater  than  the 
sum  of  A  and  B. 

II.  If  it  had  been  equal  to  the  sum  of  A  and  B. 

III.  If  it  had  been  equal  to  the  difference  of  A  and  B. 

IV.  If  it  had  been  less  than  the  difference  of  A  and  B. 
(See  Theorem  229.) 

625.  It  is  required  to  find  the  point  which  is  equally  distant 
from  two  given  points. 

Post.  Let  A  and  B  be  the  two  given  points. 

We  are  to  find  a  point  which  is  the  same  distance  from  A 
as  from  B. 


136 


PLANE  GEOMETRY. 


It  is  evident  that  whether  there  be  more  or  not,  there  must 
at  least  be  one  midway  between  A  and  B ;  so  join  AB. 

With  A  and  B  as  centres,  construct,  with  the  same  radius, 
two  circles  which  shall  intersect.  How  can  you  tell  whether  or 
not  they  will  intersect  ? 

K 


\ 


\ 


/ 


N 

Connect  the  points  of  intersection,  as  D  and  H.    Then  DH 
sustains  what  relation  to  the  two  circles  ? 

The  line  AB  sustains  what  relation  to  the  two  circles  ? 
Then  what  relation,  as  regards  their  position,  between  AB 


How  is  the  point  D  situated  with  reference  to  points  A  and  B  ? 

How  is  the  point  .ff  situated  with  reference  to  the  same  points? 

Then  what  must  be  the  relation  of  AC  and  CB  as  regards 
magnitude  ?  (Consult  Theorem  82  or  230.) 

Suppose,  now,  DH  be  indefinitely  extended,  and  any  point 
in  it  selected  at  random.  How  will  this  point  be  situated  with 
reference  to  the  points  A  and  B  ?  Why  ? 

What  name,  then,  shall  we  give  to  the  line  NK?  Why  ?  Q.E.F. 

626.    It  is  required  to  bisect  a  given  straight  line. 
Sug.   Employ  method  similar  to  the  previous  one. 


PLANE  GEOMETRY.  137 

627.   It  is  required  to  construct  a  perpendicular  to  a  given 
straight  line  which  shall  pass  through  a  given  point  in  that  line. 


Post.   Let  AB  be  the  given  line,  and  C  the  given  point. 

We  are  required  to  construct  a  perpendicular  to  AB  passing 
through  point  C. 

Sug.  Lay  off  equal  distances  each  side  of  (7,  as  CD  and 
CB ;  then  use  Problem  625. 

627  (a).  It  is  required  to  construct  a  perpendicular  to  a 
line  at  one  extremity. 

Sug.  Extend  the  line ;  then  use  previous  problem.  In  case 
the  extension  should  not  be  possible  or  convenient,  use  the 
following. 

Select  any  point  at  random,  as  (7, 

making  sure  that  it  lies  between  A  ^  ^N 

and  B.  Then,  with  CB  as  a  radius, 
construct  the  circle  or  arc  KBN. 
Through  D,  point  where  this  circle 
intersects  the  given  line,  and  (7,  draw 
the  straight  line  DH,  the  point  H 

being  where  this  line  intersects  the          "^^s,  ^  B 

arc  KBN.     Join  HB. 

Then  HB  is  the  perpendicular  required.     Why  ?  Q.E.F. 

628.  It  is  required  to  construct  a  perpendicular  to  a  given 
line  from  a  given  point  without  the  line. 

C' 


JD  s 


Let  ^15  be  the  given  line,  and  C  the  given  point. 


138  PLANE  GEOMETRY. 

Sug.  With.  C  as  a  centre,  construct  an  arc  which  shall  inter- 
sect AB. 

Point  C  is  how  situated  with  reference  to  the  two  points  of 
intersection  D  and  H  ? 

Can  you  find  another  point  the  same  distance  as  C  from 
those  two  points  ? 

If  that  point  and  point  C  be  joined  by  a  straight  line,  how 
will  that  line  be  situated  with  reference  to  DH  ?  (See 
Theorem  82.) 

628  (a).    It  is  required  to  bisect  a  given  arc. 
Sug.   Connect  the  extremities  of  the  given  arc;  then  use 
Problem  626,  and  for  proof  consult  Theorem  199. 

629.  It  is  required  to  bisect  a  given  angle. 
Sug.   Use  Problem  628  (a)  and  Theorem  192. 

630.  At  a  given  point  in  a  given  line,  it  is  required  to  con- 
struct an  angle  equal  to  a  given  angle. 

Sug.   Consult  191  and  192. 

631.  It  is  required  to  draw  through  a  given  point  a  line 
parallel  to  a  given  line. 

Sug.   Consult  87  and  84,  and  Problem  630. 

632.  Two  angles  of  a  triangle  being  given,  it  is  required  to 
construct  the  third  angle. 

Sug.    Consult  Theorems  58  and  98,  and  Problem  630. 

633.  Having  given  two  sides  of  a  triangle  and  their  included 
angle,  it  is  required  to  construct  the  triangle. 

634.  Having  given  two  angles  of  a  triangle  and  the  side 
joining  their  vertices,  it  is  required  to  construct  the  triangle. 

635.  Having  given  the  three  sides  of  a  triangle,  it  is  re- 
quired to  construct  the  triangle. 

636.  It  is  required  to  construct  the  locus  of  the  point  which 
is  a  given  distance  from  a  given  straight  line. 


PLANE  GEOMETRY.  139 

637.  It  is  required  to  construct  the  locus  of  the  point  which 
is  a  given  distance  from  a  given  circumference. 

638.  It  is  required  to  construct  the  locus  of  a  point  which 
is  equally  distant  from  two  given  parallel  lines. 

639.  It  is  required  to  construct  the  locus  of  the  point  which 
is  equally  distant  from  two   non-parallel  lines  in  the  same 
plane. 

640.  It  is  required  to  construct  the  locus  of  the  point  which 
is  equally  distant  from  the  circumferences  of  two  equal  circles. 

641.  Having  given  the  hypothenuse  of  a  right  triangle,  it 
is  required  to  construct  the  locus  of  the  vertex  of  the  right 
angle. 

Bug.   Consult  Theorem  236  (6). 

642.  It  is  required  to  find  in  a  given  line,  as  AB,  a  point 
which  is  equally  distant  from  two  given  points,  as  C  and  D. 


•D 

Sug.   Consult  Problem  625. 

643.  It  is  required  to  find  a  point  which  is  equally  distant 
from  three  given  points. 

Sug.   Join  the  points,  then  consult  Problem  625. 

644.  Through  a   given   point   without   a  given  line,  it  is 
required  to  draw  a  line  which  shall  make  an  angle  with  the 
given  line  equal  to  a  given  angle. 

645.  It  is  required  to  construct  the  triangle,  having  given 
the  base,  the  vertical  angle,  and  one  of  the  other  angles. 

646.  It  is  required  to  construct  the  triangle,  having  given 
two  sides  and  an  angle  opposite  one  of  them. 

647.  It  is  required  to  find  the  centre  of  a  given  circumfer- 
ence or  arc. 


140  PLANE  GEOMETRY. 

648.  It  is  required  to  construct  the  circumference,  having 
given  three  points  in  it. 

649.  It  is  required  to  construct  a  circumference  which  shall 
pass  through  the  vertices  of  a  given  triangle. 

650.  It  is  required  to  find  the  locus  of  the  centre  of  the 
circumference  which  shall  pass  through  two  given  points. 

651.  With  a  given  radius,  it  is  required  to  construct  the 
circle  which  shall  pass  through  two  given  points. 

652.  It  is  required  to  construct  the  isosceles  triangle,  having 
given  the  base  and  the  vertical  angle. 

653.  It  is  required  to  construct  a  circumference  which  shall 
be  a  given  distance  from  three  given  points. 

654.  It  is  required  to  construct  a  circle  which  shall  have 
its  centre  in  a  given  straight  line  and  circumference  passing 
through  two  given  points. 

655.  It  is  required  to  find  a  point  which  shall  be  equally 
distant  from  two  given  points,  and  at  a  given  distance  from  a 
third  given  point. 

656.  It  is  required  to  construct  the  equilateral  triangle,  hav- 
ing given  one  side. 

657.  It  is  required  to  trisect  a  right  angle. 

658.  It  is  required  to  find  a  point  which  shall  be  equally 
distant  from  two  given  points,  and  also  equally  distant  from 
two  given  parallel  lines. 

659.  It  is  required  to  find  a  point  which  shall  be  equally 
distant  from  two  given  points,  and  also  equally  distant  from 
two  given  non-parallel  lines  in  the  same  plane. 

660.  It  is  required  to  find  a  point  which  shall  be  equally 
distant  from  two  given  parallel  lines,  and  also  equally  distant 
from  two  non-parallel  lines  in  the  same  plane. 


PLANE  GEOMETRY.  141 

661.  It  is  required  to  construct 

I.  an  angle  of  45° ;  VI.      an  angle  of  75° ; 

II.  an  angle  of  60° ;  VII.    an  angle  of  22°  30' ; 

III.  an  angle  of  30°  ;  VIII.  an  angle  of  52°  30' ; 

IV.  an  angle  of  15°  ;  IX.      an  angle  of  135° ; 

V.  an  angle  of  105° ;  X.        an  angle  of  165°. 

662.  It    is    required   to    find   a  point   in    one    side  of  a 
triangle  which  shall  be  equally  distant  from  the  other  two 

sides. 

663.  It  is  required  to  find  a  point  which  shall  be  equally 
distant  from  two  non-parallel  lines  in  the  same  plane,  and  at 
a  given  distance  from  a  given  point. 

664.  It  is  required  to  construct  the  right  triangle,  having 
given  the  two  legs. 

665.  It  is  required  to  construct  the  right  triangle,  having 
given  the  hypothenuse  and  one  acute  angle. 

666.  It  is  required  to  construct  the  right  triangle,  having 
given  one  leg  and  adjacent  acute  angle. 

667.  It  is  required  to  construct  the  right  triangle,  having 
given  one  leg  and  the  acute  angle  opposite. 

668.  It  is  required  to  construct  the  right  triangle,  having 
given  the  hypothenuse  and  one  leg. 

669.  It  is  required  to  construct  a  tangent  to  a  given  circle, 
having  given  the  point  of  contact. 

670.  It  is  required  to  construct  a  tangent  to  a  given  circle, 
which  shall  pass  through  a  given  point  outside  the  circle. 

Sug.  Connect  the  centre  of  the  given  circle  with  the  given 
exterior  point.  On  this  line  as  a  diameter  construct  a  circle, 
and  join  the  points  of  its  intersection  of  the  given  circle  with 
extremities  of  the  diameter. 


142 


PLANE  GEOMETRY. 


671.  It  is  required  to  construct  the  parallelogram,  having 
given  two  sides  and  their  included  angle. 

672.  It  is  required  to  construct  a  circle  within  a  given  tri- 
angle, so  that  the  sides  of  the  triangle  shall  be  tangents  of  the 
circle. 

Sug.   Consult  Theorems  117  and  125. 

673.  It  is  required  to  construct  the  circle,  having  given  a 
chord  and  angle  made  by  the  chord  and  a  tangent. 


Let  AB  be  the  given  chord,  and  C  the  angle  made  by  the 
chord  and  tangent.  Then  one  extremity  of  the  chord  must  be 
the  point  of  contact. 

Take  D.H=AB.  Construct  angle  DHN  equal  to  angle  C. 
Then  H  is  the  point  of  contact,  and  NH  the  tangent. 

Find  the  locus  of  the  centre  of  the  circumference  passing 
through  V  and  H.  (Problem  650.) 

Then  consult  Theorem  202. 

Hence  T  must  be  the  centre  of  the  circle  required. 

If  from  any  point  in  the  arc  DOH  (call  the  point  Q)  lines 
be  drawn  to  D  and  ff,  how  will  the  angle  Q  compare  in  magni- 
tude with  the  angle  DHN? 


PLANE  GEOMETRY.  143 

What  is  formed  by  the  lines  DH,  DQ,  and  HQ  ? 
If  we  call  DH  the  base,  what  would  you  call  the  angle  Q  ? 
What  the  point  Q  ? 

What  might  the  arc  DQH  be  named,  then  ?  Q.E.F. 

674.  Having  given  the  base  and  vertical  angle  of  a  triangle, 
it  is  required  to  construct  the  locus  of  its  vertex. 

Sug.    Consult  the  previous  problem. 

675.  It  is  required  to  construct  the  triangle,  having  given 
the  base,  the  vertical  angle,  and  the  altitude. 

Sug.    Use  previous  problem. 

676.  It  is  required  to  construct  the  triangle,  having  given 
the  base,  the  vertical  angle,  and  the  median. 

OPTIONAL  PKOBLEMS  FOB  ADVANCE  WOEK. 

677.  It  is  required  to  construct  the  triangle,  having  given 
the  base,  vertical  angle,  and  perpendicular  from  one  extremity 
of  the  base  to  the  opposite  side. 

678.  It  is  required  to  construct  the  isosceles  triangle,  having 
given  the  altitude  and  one  of  the  equal  angles. 

679.  It  is  required  to  construct  the  chord  in  a  given  circle, 
having  given  the  middle  point  of  the  chord. 

680.  It  is  required  to  construct  a  circle  whose  circumference 
shall  pass  through  the  vertices  of  a  given  rectangle. 

681.  It  is  required  to  construct  a  tangent  to  a  given  circle 
which  shall  be  parallel  to  a  given  straight  line. 

682.  It  is  required  to  construct  a  tangent  to  a  given  circle 
which  shall  be  perpendicular  to  a  given  straight  line. 

683.  It  is  required  to  construct  a  tangent  to  a  given  circle 
which  shall  make  a  given  angle  with  a  given  straight  line. 

684.  It  is  required  to  construct  the  isosceles  triangle,  having 
given  the  vertical  angle  and  a  point  in  the  base,  in  position. 


144  PLANE  GEOMETRY. 

685.  It  is  required  to  construct  the  triangle,  having  given 
the  altitude,  the  base,  and  an  adjacent  angle. 

686.  It  is  required  to  construct  the  triangle,  having  given 
the  altitude,  the  base,  and  an  adjacent  side. 

687.  It  is  required  to  construct  a  rhombus,  having  given  its 
base  and  altitude. 

688.  It  is  required  to  construct  the  triangle,  having  given 
the  altitude  and  the  sides  which  include  the  vertical  angle. 

689.  It  is  required  to  construct  the  triangle,  having  given 
the  altitude  and  angles  adjacent  to  the  base. 

690.  It  is  required  to  construct  an  isosceles  triangle  which 
shall  have  its  vertical  angle  twice  the  sum  of  its  other  two 
angles. 

691.  It  is  required  to  construct  the  square,  having  given  its 
diagonal. 

692.  It  is  required  to  construct  the  right  triangle,  having 
given  the  hypothenuse  and  perpendicular  from  the  vertex  of 
the  right  angle  to  the  hypothenuse. 

693.  It  is  required  to  construct  the  locus  of  the  centre  of 
the  circle  of  given  radius  tangent  to  a  given  straight  line. 

694.  It  is  required  to  construct  the  circle  of  given  radius 
which  shall  be  tangent  to  two  given  non-parallel  lines. 

695.  It  is  required  to  construct  the  circle  of  given  radius 
which  shall  be  tangent  to  a  given  straight  line,  and  whose  cen- 
tre shall  be  in  a  given  line  not  parallel  to  the  former. 

696.  It  is  required  to  construct  the  circle  which  shall  be 
tangent  to  a  given  line,  and  whose  circumference  shall  pass 
through  a  given  point. 

697.  It  is  required  to  find  the  locus  of  the  centre  of  the 
circle  of  given  radius  which  shall  be  tangent  externally  to  a 
given  circle. 


PLANE  GEOMETRY  145 

It  is  required  to  construct  the  locus  of  the  centre  of 
the  circle  of  given  radius  which  shall ,  be  tangent  internally  to 
a  given  circle. 

699.  It  is  required  to  construct  a  circle  which  shall  be  tan- 
gent to  a  given  line  and  a  given  circle. 

700.  It  is  required  to  construct  a  circle  which  shall  be  tan- 
gent to  two  given  circles. 

701.  It  is  required  to  construct  a  circle  which  shall  cut 
three  equal  chords  of  given  length  from  three   given  non- 
parallel  lines. 

702.  It  is  required  to  construct  in  a  given  circle  a  chord  of 
given  length  passing  through  a  given  point. 

703.  It  is  required  to  construct  in  a  given  circle  a  chord  of 
given  length  and  parallel  to  a  given  straight  line. 

704.  It  is  required  to  construct  a  line  of  given  length  pass- 
ing through  a  given  point  between  two  given  parallel  lines. 

705.  It  is  required  to  construct  a  line  of  given  length  between 
two  given  non-parallel  lines,  and  which  shall  be  parallel  to  a 
given  line. 

706.  It  is  required  to  construct  a  line  of  given  length  between 
two  non-parallel  lines,  and  which  shall  pass  through  a  given 
point. 

707.  It  is  required  to  construct  the  right  triangle,  having 
given  the  hypothenuse  and  radius  of  the  inscribed  circle. 

708.  It  is  required  to  construct  the  right  triangle,  having 
given  the  radius  of  the  inscribed  circle  and  one  acute  angle. 

709.  It  is  required  to  construct  the  triangle,  having  given 
in  position  the  middle  points  of  its  sides. 

710.  It  is  required  to  construct  the  triangle,  having  given 
the  base,  vertical  angle,  and  radius  of  the  circumscribing  circle. 


146  PLANE  GEOMETRY. 

711.  It  is  required  to  construct  the  isosceles  triangle,  having 
given  the  base  and  radius  of  the  inscribed  circle. 

712.  It  is  required  to  construct  a  straight  line  which  shall 
pass  through  a  given  point  and  make  equal  angles  with  two 
given  lines. 

713.  It  is  required  to  find  a  point  in  a  given  secant  to  a 
given  circle  such  that  the  tangent  to  the  circle  from  that 
point  shall  be  of  given  length. 

714.  It  is  required  to  construct  the  right  triangle,  having 
given  one  leg  and  radius  of  the  inscribed  circle. 

715.  It  is  required  to  construct  the  right  triangle,  having 
given  the  median  and  altitude  from  the  vertex  of  the  right 
angle  to  the  hypothenuse. 

716.  It  is  required  to  find  the  locus  of  the  centre  of  the 
chord  which  passes  through  a  given  point  in  a  given  circle. 

717.  It  is  required  to  construct  the  triangle,  having  given 
the  base,  vertical  angle,  and  sum  of  the  other  two  sides. 

718.  It  is  required   to   construct  a  circle  which   shall   be 
tangent  to  two  given  lines  and  at  given  point  of  contact  in  one. 

719.  It  is  required  to  inscribe  a  circle  in  a  given  sector. 

720.  It  is  required  to  construct  a  common  tangent  to  two 
given  circles. 

Sug.  Make  five  cases  according  to  relative  position  of  the 
circles.  (See  Theorem  229.) 

721.  It  is  required  to  inscribe  a  square  in  a  given  rhombus. 

722.  Having  given  two  intersecting  circles,  it  is  required  to 
draw  a  line  through  one  of  the  points  of  intersection,  so  that 
the  two  intercepted  chords  shall  be  equal. 

Sug.  Join  centres.  Draw  from  point  of  intersection  to 
middle  of  that  line.  From  centres  draw  radii  parallel  to  latter 
line.  Through  point  of  intersection  draw  line  perpendicular  to 
these  radii. 


PLANE  GEOMETRY.  147 

723.  It  is  required  to  construct  an  equilateral  triangle  having 
its  vertices  in  three  given  parallel  lines. 

724.  It  is  required  to  construct  a  tangent  to  a  given  circle 
with  two  given  parallel  secants  so  that  the  point  of  contact 
shall  bisect  the  part  between  the  secants. 

725.  It  is  required  to  construct  three  equal  circles  which 
shall  be  tangent  to   each  other  and  also  to  a  given  circle 
externally. 

726.  It  is  required  to  construct  three  equal  circles  which 
shall  be   tangent  to  each  other  and  also  to  a  given  circle 
internally. 

727.  It  is  required  to  construct  three  equal  circles  which 
shall  be  tangent  to  each  other  and  also  to  the  sides  of  an 
equilateral  triangle. 

728.  It   is   required  to   construct  a  semicircle   having  its 
diameter  in  one  of  the  sides  of  a  given  triangle  and  tangent 
to  the  other  two  sides. 

729.  It  is  required  to  construct  a  triangle,  having  given  the 
radius  of  the  inscribed  circle  and  two  sides. 

730.  It  is  required  to  find  a  point  in  a  given  line  such  that 
lines  to  that  point  from  two  given  points  without  the  line 
make  equal  angles  with  the  line. 

731.  It  is  required  to  construct  the  triangle,  having  given 
the  perimeter,  altitude,  and  vertical  angle. 

EEQUIRED  PROBLEMS  OF  CONSTRUCTION. 
It  is  required, 

732.  To  divide  a  given  line  into  any  number  of  equal  parts. 
Sug.    From  one  extremity  of  the  given  line  construct  a  line 

of  indefinite  length,  making  any  convenient  angle  with  the 
given  line.  Then,  with  any  convenient  unit  of  length  assumed 
as  a  unit  of  measure,  beginning  at  the  vertex,  lay  off  on  the 


148  PLANE  GEOMETRY. 

indefinite  line  this  unit  as  many  times  as  it  is  required  to 
divide  the  given  line  into  parts.     Then  consult  316. 

733.  To  divide  a  given  line  into  parts  proportional  to  any 
number  of  given  lines. 

Sug.  Place  the  given  parts  so  as  to  form  one  straight  line, 
making  any  convenient  angle  with  the  line  to  be  divided. 
Then  consult  313  and  319. 

734.  To  construct  a  line  that  shall  be  a  fourth  proportional 
to  three  given  lines. 

Sug.  Consult  Ratio  and  Proportion,  277,  Theorem  313,  and 
previous  problem. 

735.  To  construct  a  line  that  shall  be  a  third  proportional 
to  two  given  lines. 

Sug.    Consult  Katio  and  Proportion,  279. 

What  must  one  of  the  given  lines  be  if  those  two  and  one 
other  are  to  form  a  proportion?  Consider  that  in  the  con- 
struction. 

736.  To  construct  a  mean  proportional  between  two  given 
lines. 

Sug.   Consult  Theorem  367. 

737.  Given  a  polygon  and  an  homologous  side  of  another 
similar  polygon,  to  construct  the  latter. 

Sug.   Consult  Theorems  356  and  357. 

738.  To  inscribe  in  a  given  circle  a  triangle  similar  to  a 
given  triangle. 

739.  To  circumscribe  about  a  given  circle  a  triangle  similar 
to  a  given  triangle. 

740.  To  construct  a  square  which  shall  be  equivalent  to  the 
sum  of  two  given  squares. 

Sug.   Consult  Theorem  360. 

741.  To  construct  a  square  which  ktiall  be  equivalent  to  the 
sum  of  three  or  more  given  squar.es.. 


PLANE  GEOMETRY.  149 

Sug.  Construct  a  square  equivalent  to  two  of  the  given 
squares,  then  one  equivalent  to  that  and  one  other  of  the 
given  squares,  and  so  on. 

742.  To  construct  a  square  which  shall  be  equivalent  to 
the  difference  of  two  given  squares. 

743.  To  construct  a  square  equivalent  to  a  given  rectangle. 
JSug.   If  x  and  y  are  the  base  and  altitude  of  a  rectangle, 

and  n  one  side  of  an  equivalent  square,  then 

xy  =  n2. 

Whence  x :  n  : :  n  :  y ;          (See  Theorem  294.) 

or  n  is  a  mean  proportional  between  x  and  y. 

Hence  consult  Problem  736. 

744.  To  construct  a  square  which  shall  be  equivalent  to  a 
given  parallelogram. 

745.  To  construct  a  square  which  shall  be  equivalent  to  a 
given  triangle. 

746.  To  construct  a  triangle  which  shall  be  equivalent  to  a 
given  polygon  of  more  than  three  sides. 

Let  ABCDH  be  a  polygon 
of  n  sides. 

Extend  one  of  the  sides,  as 
AB,  and  construct  the  diag- 
onal DB.  Through  vertex  C 
draw  CK  parallel  to  DB,  and 
join  DK. 

Considering  DB  the  com- 
mon base  of  the  two  triangles  DCB  and  DKB,  what  is  the 
relation  between  the  areas  of  those  two  triangles  ?     Why  ? 

How  does  the  area  of  the  polygon  DKAH,  then,  compare 
with  that  of  DCBAH?  Why? 

How  many  sides  has  the  former  as  compared  with  the 
latter  ? 

Proceed  in  the  same  way  with  the  polygon  DKAB. 


150 


PLANE  GEOMETRY. 


746  (a).  To  construct  a  square  equivalent  to  any  given  polygon. 
Sug.    Use  Problem  746,  then  745. 

746  (6).  To  construct  a  square  equivalent  to  the  sum  of  any 
number  of  given  polygons. 

746  (c).  To  construct  a  rectangle  which  shall  be  equivalent 
to  a  given  square  and  the  sum  of  whose  base  and  altitude 
shall  be  equal  to  a  given  line. 

Sug.  Upon  the  given  line  as  a  diameter  construct  a  circle. 
Construct  a  line  parallel  to  the  diameter,  distant  from  it  one 
side  of  the  given  square.  Then  consult  Theorem  367. 

746  (d)  .  To  construct  a  rectangle  which  shall  be  equivalent 
to  a  given  square  and  the  difference  of  whose  base  and  altitude 
shall  be  equal  to  a  given  line. 

Sug.  Proceed  as  in  746  (c),  then  at  extremity  of  the  diam- 
eter construct  a  tangent  equal  to  one  side  of  given  square,  and 
from  other  extremity  of  this  tangent  construct  a  secant 
through  centre  of  circle.  (Consult  Theorem  370.) 

747.  To  construct  a  right  triangle  which  shall  be  equivalent 
to  a  given  triangle  and  its  hypothenuse  equal  to  a  given  line. 

748.  To  divide  a  given  line  into  extreme  and  mean  ratio. 
(See  280.) 


Post.   Let  AB  be  the  given  line.     At  one  extremity  erect 
a  perpendicular  CB  equal  to  one-half  AB.     With  C  as  centre 


PLANE  GEOMETRY. 


151 


Dem.   Since  CjB=—    DK=AB. 


and  CB  as  a  radius  construct  the  circle  DBN.  Construct  the 
secant  AK  passing  through  the  centre  C.  With  A  as  a  centre 
and  radius  AD  (point  D  being  point  of  intersection  of  secant 
and  circumference)  construct  the  arc  DH.  Then  the  line  AB 
will  be  divided  in  extreme  and  mean  ratio  at  H. 

2 

AD  :  AB  :  :  AB  :  AK.  Why  ? 

AD  :  AB  -  AD  :  :  AB  :  AK-  AB.  Why  ? 

AH  :  AB  -  AH:  :  AB  :  AK-DK.  Why  ? 

AH:HB::AB:AD.  Why? 

AH:HB::AB:AH',  Why? 

or  HB:AH::AH:  AB.  Why  ? 

Hence  the  line  AB  is  divided  in  extreme  and  mean  ratio. 

Q.E.F. 

749.   To  inscribe  a  regular  decagon  in  a  given  circle. 

Post.  Let  ABH  be  the  given 
circle,  and  A  C  one  of  its  radii. 

Consult  Theorem  546,  then  use 
Problem  748. 

Cons.  With  A  as  a  centre  and 
D  C  as  a  radius,  construct  chord 
AB.  Join  BC  and  BD. 

Dem.    AC:AB::AB:AD. 
Why? 

Hence  the  two  AABD  and 
ABO  are  how  related  ?  (See  Theorem  350.) 

What  kind  of  a  triangle  is  ^OB  ? 

What  kind  of  a  triangle  must  ABD  be,  then  ? 

What  relation,  then,  between  AB  and  DB  ?    What  between 


What  relation,  then,  between  the  A  CAB  and  <3R4  ?    Be- 
tween the  A  DAB  and   £AD?     Between   the   A  BDA  and 
?    Why  ?    Between  the  A  DBC  and  DCB  ? 


152  PLANE  GEOMETRY. 

What  relation  does  the  Z  BDA  bear  to  the  sum  of  DCB  and 
DBG?     Why? 

What  relation  does  Z  DAB  bear  to  Z  C,  then  ? 

Hence,  what  relation  does  Z  AB(7  bear  to  Z  0  ? 

Compare  now  Z  DAB  +  Z  ABC  with  the  Z  (7,  and  finally 
compare  Z  1MB  +  Z  ABC  +  Z  0  with  the  Z  (7. 

If  the  pupil  has  answered  the  above  questions  correctly,  he 
will  now  have  the  equation, 

Z  DAB  +  Z  JLBC  +  Z  <7  =  5  Z  G. 

What  is  the  value  of  the  first  member  of  the  above  equa- 
tion ?     Why  ? 

Then  5ZC=2rt.Zs; 

or  10  Z  C  =  4  rt.  A. 

Whence  Z  <7  =  TV  of  4  rt.  A 

Hence  the  arc  ^..B  is  what  part  of  the  circumference  ? 

.-.  AB  is  the  side  of  a  regular  inscribed  decagon.  Q.E.F. 

750.  To  inscribe  a  square  in  a  given  circle. 
tSug.    Consult  Theorem  513. 

751.  To  inscribe  a  regular  hexagon  in  a  given  circle. 
Sug.   Consult  Theorem  480. 

752.  To  inscribe  a  regular  pentedecagon  in  a  given  circle. 
Sug.   Find  the   difference   between  a  central  angle  of  the 

regular  decagon  and  that  of  a  regular  hexagon. 

753.  To  inscribe  in  a  given  circle 

I.  a  regular  trigon  ; 

II.  a  regular  pentagon  ; 

III.  a  regular  octagon  ; 

IV.  a  regular  dodecagon ; 

V.  a  regular  polygon  of  sixteen  sides ; 

VI.  a  regular  polygon  of  twenty  sides. 

754.  To  circumscribe  around  a  given  circle  all  the  above- 
mentioned  regular  polygons. 


PLANE  GEOMETRY. 


153 


755.  To  inscribe  in  a  given  circle  a  regular  polygon  similar 
to  a  given  regular  polygon. 

Sug.   Construct  a  central  angle  equal  to  that  of  the  given 
polygon,  etc. 

756.  To    circumscribe   a  circle    about    any  •  given    regular 
polygon. 

757.  Upon  a  given  line  as  a  base,  to  construct  a  rectangle 
equivalent  to  a  given  rectangle. 

758.  To  construct  a  square  whose  ratio  to  a  given  square 
shall  be  the  same  as  that  of  two  given  lines. 


Post.  Let  Q  be  the  given  square,  and  I  and  p  the  two  given 
lines. 

We  are  required  to  construct  a  square  (Q')  so  that 

Area  Q  :  Area  Q' : :  I :  p. 

Cons.  Place  the  two  given  lines  so  as  to  form  one  straight 
line,  &s\ADB. 

On  this  as  a  diameter  construct  a  semicircle,  and  at  Z>  erect 
the  perpendicular  DK.  Join  AK  and  BK. 

Make  KN  equal  to  one  side  of  the  given  square,  as  ST. 

Draw  NH  parallel  to  AB.  Then  KH  is  the  side  of  the 
required  square. 

Dem.  ZK2 :  BK2  ::AD:  DB.      (Theorem  360,  IV.) 

Or,  AK2:BK2::l:p. 

Again,  NK  :  HK  : :  AK  :  BK.  Why  ? 

Hence  NK2 :  HK2 : :  AK2 :  BK2. 

.'.  NK2  :HK2::l:p.  Why? 


154  PLANE  GEOMETRY. 

Hence  the  square  constructed  on  HK  as  a  side  is  the  re- 
quired square.  Q.E.F. 

759.    To  construct  a  polygon  similar  to  a  given  polygon,  the 
ratio  of  whose  areas  shall  be  that  of  two  given  lines. 


C 

Post.  Let  ABCDH  be  the  given  polygon  (P),  and  p  and  q 
the  two  given  lines. 

We  are  required  to  construct  a  polygon  (P')  similar  to  P, 
so  that  Area  p  .  Area  P,  ::p:q 

Cons.  Upon  any  side  of  the  polygon  P,  as  AB,  construct  a 
square. 

Then,  by  the  previous  problem,  find  the  side  of  a  square 
whose  ratio  to  that  of  the  square  on  AB  shall  be  that  of  the 
two  lines  p  and  q. 

Upon  this  line  construct  a  polygon  similar  to  polygon  P. 
This  will  be  the  polygon  required. 

Dem.   This  will  be  left  for  the  pupil. 

759 (a).  To  construct  a  polygon  similar  to  one  of  two  given 
polygons  and  equivalent  to  the  other. 


Q 


Post.   Let  P  and  Q  be  the  two  given  polygons.      We  are 


PLANE  GEOMETRY.  155 

required  to  construct  a  polygon  similar  to  P  and  equivalent 
to  Q. 

Cons.  Construct  squares  equivalent  to  each  of  the  polygons 
P  and  Q. 

Then  find  a  fourth  proportional  to  the  sides  of  these  squares 
and  any  side  of  the  polygon  P  selected  at  random,  as  AB. 
Upon  this  fourth  proportional  as  an  homologous  side  construct 
a  polygon  similar  to  P.  Then  this  polygon  will  be  similar  to 
P  and  equivalent  to  Q,  and  is  therefore  the  polygon  required. 

Dem.   This  is  also  left  for  the  pupil. 

760.  To  construct  upon  a  given  line,  as  one  side, 

I.  a  regular  trigon  ;  V.       a  regular  pentagon ; 

II.  a  regular  tetragon ;  YI.      a  regular  decagon. 

III.  a  regular  hexagon ;  VII.     a  regular  dodecagon  ; 

IV.  a  regular  octagon ;  VIII.  a  regular  pentedecagon. 

761.  To  construct  a  regular  hexagon,  given  one  of  its  shorter 
diagonals. 

762.  To   construct   a  regular  pentagon,   given   one  of  its 
diagonals. 

763.  To  construct  a  circle  equivalent  to  the  sum  of  two 
given  circles. 

764.  To  construct  a  circumference  equal  to  the  sum  of  two 
given  circumferences. 

765.  To  divide  a  given  circle  by  a  concentric  circumference 
into  two  equal  parts. 

MISCELLANEOUS  PLANE  PROBLEMS  FOR  ADVANCE  WORK. 

I.     TRIANGLES. 
It  is  required  to  construct  the  triangle,  having  given, 

766.  Its  base,  vertical  angle,  and  difference  of  the  other  two 
sides. 


156  PLANE  GEOMETRY. 

767.  Its  base,  vertical  angle,  and  a  square  which  is  equal  to 
the  sum  of  the  squares  upon  the  other  two  sides. 

768.  Its  base,  vertical  angle,  and  a  square  which  is  equiva- 
lent to  the  difference  of  the  squares  upon  the  other  two  sides. 

769.  Its  base,  vertical  angle,  and  sum  of  its  altitude  and  the 
two  remaining  sides. 

770.  Its  base,  vertical  angle,  and  the  sum  of  its  altitude 
and  difference  of  the  other  two  sides. 

771.  Its  base,  vertical  angle,  and  difference  between  its  alti- 
tude and  sum  of  its  other  two  sides. 

772.  Its  base,  vertical  angle,  and  difference  between  its  alti- 
tude and  difference  of  the  other  two  sides. 

773.  Its  base,  vertical  angle,  and  ratio  of  its  altitude  to  the 
sum  of  its  other  two  sides. 

774.  Its  base,  vertical  angle,  and  ratio  of  its  altitude  to  the 
difference  of  its  other  two  sides. 

775.  Its  base,  altitude,  and  sum  of  its  other  two  sides. 

776.  Its  base,  altitude,  and  difference  of  its  other  two  sides. 

777.  Its  base,  altitude,  and  ratio  of  the  other  two  sides. 

778.  Its  base,  altitude,  and  a  square  equivalent  to  the  rec- 
tangle of  the  other  two  sides. 

779.  Its  base,  altitude,  and  a  square  which  is  equivalent  to 
the  sum  of  the  squares  upon  the  other  two  sides. 

780.  Its  base,  altitude,  and  a  square  which  is  equivalent  to 
the  difference  of  the  squares  upon  the  other  two  sides. 

781.  Its  vertical  angle,  sum  of  base  and  altitude,  and  sum 
of  the  other  two  sides. 

782.  Its  vertical  angle,  sum  of  base  and  altitude,  and  differ- 
ence of  its  other  two  sides. 


PLANE  GEOMETRY.  157 

783.  Its  vertical  angle,  sum  of  base  and  altitude,  and  ratio 
of  the  other  two  sides. 

784.  Its  vertical  angle,  sum  of  base  and  altitude,  and  a 
square  equivalent  to  the  rectangle  of  its  other  two  sides. 

785.  Its  vertical  angle,  sum  of  base  and  altitude,  and  sum 
of  the  three  sides. 

786.  Its  vertical  angle,  sum  of  base  and  altitude,  and  differ- 
ence between  the  base  and  sum  of  the  other  two  sides. 

787.  Its  vertical  angle,  sum  of  base  and  altitude,  and  differ- 
ence between  the  base  and  difference  of  its  other  two  sides. 

788.  Its  vertical  angle,  sum  of  base  and  altitude,  and  the 
ratio  of  the  base  to  the  sum  of  the  other  two  sides. 

789.  Its  vertical  angle,  sum  of  base  and  altitude,  and  the 
ratio  of  the  base  to  the  difference  of  its  other  two  sides. 

790.  Its  vertical  angle,  altitude,  and  the  square  equivalent 
to  the  sum  of  the  squares  of  the  sides  which  include  the  ver- 
tical angle. 

791.  Its  vertical  angle,  altitude,  and  radius  of  the  circum- 
scribing circle. 

792.  Its  vertical  angle,  radius  of  the  inscribed  circle,  and 
perimeter. 

793.  Its  vertical  angle,  radius  of  the  inscribed  circle,  and 
ratio  of  the  sides  including  the  vertical  angle. 

794.  Its  vertical  angle,  radius  of  the  inscribed  circle,  and 
a  square  equivalent  to  the  rectangle  of  the  sum  of  the  two 
sides  including  the  vertical  angle  and  the  base. 

795.  Its  vertical  angle,  radius  of  the  inscribed  circle,  and 
a  square  whose  area  is  equal  to  the  difference  between  the 
sum  of  the  squares  of  the  sides  including  the  vertical  angle 
and  the  square  of  the  base. 

796.  Its  base,  meclian,  and  sum  of  the  other  two  sides. 


158  PLANE  GEOMETRY. 

797.  Its  base,  median,  and  difference  of  the  other  two  sides. 

798.  Its  three  altitudes. 

799.  Its  three  medians. 

800.  Two  sides,  and  difference  of  the  angles  opposite  them. 

801.  Its  vertical  angle,  difference  of  the  angles  at  the  base, 
and  difference  of  the  other  two  sides. 

802.  Difference  of  the  angles  at  the  base,  difference  of  the 
segments  of  the  base  made  bj  the  altitude,  and  sum  of  the 
other  two  sides. 

803.  It  is  required  to   construct   the  equilateral   triangle, 
having  given  the  three  distances  from  its  vertices  to  a  common 
point  within  the  triangle. 

804.  The  same  as  803,  but  the  common  point  without  the 
triangle. 

IL     QUADRILATERALS. 

805.  It  is  required  to  construct  a  square,  having  given 

I.  the  sum  of  its  diagonal  and  side. 

II.  the  difference  of  its  diagonal  and  side. 

It  is  required  to  construct  a  rectangle,  having  given 

806.  The  sum  of  two  adjacent  sides  and  its  diagonal. 

807.  The  difference  of  two  adjacent  sides  and  its  diagonal 

808.  One  side,  and  sum  of  diagonal  and  adjacent  side. 

809.  One  side,  and  difference  of  diagonal  and  adjacent  side. 
It  is  required  to  construct  the  rhombus,  having  given 

810.  Its  side  and  altitude. 

811.  Its  altitude  and  lesser  angle. 

812.  Its  side  and  sum  of  its  diagonals. 

813.  Its  side  and  difference  of  its  diagonals. 


PLANE  GEOMETRY.  159 

814.  Its  lesser  angle  and  sum  of  its  diagonals. 

815.  Its  lesser  angle  and  difference  between  its  longer  diago- 
nal and  altitude. 

It  is  required  to  construct  the  rhomboid,  having  given 

816.  The  longer  side,  sum  of  its  diagonals,  and  larger  angle 
made  by  the  diagonals. 

817.  Its  lesser  angle,  longer  diagonal,  and  sum  of  two  adja- 
cent sides. 

818.  Its  lesser  angle,  longer  side,  and  sum  of  its  altitude 
and  lesser  side. 

819.  Its  lesser  angle,  shorter  side,  and  difference  of  its 
longer  diagonal  and  longer  side. 

It  is  required  to  construct  an  isosceles  trapezoid,  having 
given, 

820.  One  leg,  diagonal,  and  longer  base. 

821.  Its  longer  base,  diagonal,  and  lesser  angle. 

822.  Its  diagonal,  altitude,  and  leg. 

823.  Its  longer  base,  lesser  angle,  and  sum  of  altitude  and 
leg. 

It  is  required  to  construct  the  trapezoid,  having' given 

824.  Its  longer  base,  lesser  angle  (i.e.  angle  formed  by  longer 
base  and  a  leg),  and  its  diagonals. 

825.  Its  longer  base,  one  leg,  lesser  angle,  and  altitude. 

826.  Sum  of  its  bases,  the  two  legs,  and  lesser  angle. 

827.  Difference  of  its  bases,  the  two  legs,  and  angle  formed 

by  its  diagonals. 

III.     CIRCLES. 

Having  given  a  circle,  it  is  required  to  construct 

828.  Three  equal  circles,  tangent  to  the  given  circle  exter- 
nally, and  tangent  to  each  other. 


160  2  LANE  GEOMETRY. 

829.  Three  equal  circles,  tangent  to  the  given  circle  inter- 
nally, and  tangent  to  each  other. 

830.  Four  equal  circles,  as  in  (828)  and  (829). 

831.  Five  equal  circles,  as  in  (828)  and  (829). 

832.  Six  equal  circles,  as  in  (828)  and  (289). 

833.  A  circle  tangent  to  three  given  circles. 

IV.     TRANSFORMATION  OF  FIGURES. 
It  is  required, 

834.  To   transform   a   given   triangle   into   an   equivalent 
isosceles  one  having  the  same  base. 

835.  To  transform  a  given  isosceles  triangle  into  an  equiva- 
lent equilateral  one. 


836.   To   transform   a   given   triangle    into    an   equivalent 
equilateral  one. 


837.    To  transform  a  given  triangle  into  another  equivalent 
triangle  whose  base  and  altitude  shall  be  equal. 


838.  To  transform  a  given  triangle  into  another  equivalen 
triangle,  and  similar  to  a  given  triangle. 

839.  To  transform   a  given  triangle  into  a  triangle  with 
one  angle  unchanged  and  its  opposite  side  parallel  to  a  given 
line. 

840.  To  transform  a  given  triangle  into  an  equivalent  tri- 
angle with  a  given  perimeter. 

841.  To  transform  a  triangle  into  a  trapezoid,  one  of  whose 
bases  shall  be  the  base  of  the  triangle,  and  one  of  its  adjacent 
angles  one  of  the  basal  angles  of  the  triangle. 


: 


PLANE  GEOMETRY.  161 

842.  To  transform  a  given  triangle  into  a  right  triangle  with 
given  perimeter. 

843.  To  transform  a  given  triangle  into  a  parallelogram  with 
given  base  and  altitude. 

844.  To  transform  a  parallelogram  into  a  parallelogram  with 
a  given  side. 

845.  To  transform  a  parallelogram  into  a  parallelogram  hav- 
ing a  given  angle.    • 

846.  To  transform  a  parallelogram  into  a  parallelogram  with 
given  altitude. 

To  transform  a  square  into 

847.  A  right  triangle. 

848.  An  isosceles  triangle. 

849.  An  equilateral  triangle. 

850.  A  rectangle  with  given  side. 

851.  A  rectangle  with  given  perimeter. 

852.  A  rectangle  with  given  difference  of  sides. 

853.  A  rectangle  with  given  diagonal. 
To  transform  a  rectangle  into 

854.  A  square. 

855.  An  isosceles  triangle. 

856.  An  equilateral  triangle. 

857.  A  rectangle  with  given  side. 

858.  A  rectangle  with  given  perimeter. 

859.  A  rectangle  with  given  difference  of  sides. 

860.  A  rectangle  with  given  diameter. 

It  is  required  to  construct  a  parallelogram  equivalent  to  the 

861.  Sum  of  two  given  parallelograms  of  equal  altitudes. 


162  PLANE  GEOMETRY. 

862.  Difference  of  two  given  parallelograms  of  equal  alti- 
tudes. 

863.  Sum  of  two  given  parallelograms  of  equal  bases. 

864.  Difference  of  two  given  parallelograms  of  equal  bases. 

865.  Sum  of  two  given  parallelograms. 

866.  Difference  of  two  given  parallelograms. 

It  is  required  to  transform  a  given  parallelogram  into 

867.  A  triangle. 

867  (a) .   An  isosceles  triangle. 

868.  A  right  triangle. 

868  (a) .   An  equilateral  triangle. 

869.  A  square. 

870.  A  rhombus  having  for  a  diagonal  one  side  of  the  par- 
allelogram. 

871.  A  rhombus  having  a  given  diagonal. 

872.  A  rhombus  having  a  given  side. 

873.  A  rhombus  having  a  given  altitude. 

874.  A  parallelogram  having  a  given  side  and  diagonal. 

875.  To  transform  a  rhombus  into  a  square. 

876.  To  inscribe  in  a  given  circle  a  rectangle  equivalent  to 
a  given  square. 

To  transform  a  trapezoid  into 

877.  A  triangle. 
877  (a).    A  square. 

878.  A  parallelogram  having  for  one  base  the  longer  base  of 
the  trapezoid. 

879.  An  isosceles  trapezoid. 


PLANE  GEOMETRY.  163 

To  transform  a  trapezium  into 

880.  A  triangle. 

881.  An  isosceles  triangle  with  given  base. 

882.  A  parallelogram. 

883.  A  trapezoid  with  one  side  and  the  two  adjacent  angles 
unchanged. 

V.     DIVISION  OF  FIGURES. 

884.  It  is  required  to  divide  a  given  triangle  into  any  num- 
ber of  equivalent  parts,  by  lines  drawn  from  one  vertex. 

885.  It  is  required  to  divide  a  given  triangle  into  any  num- 
ber of  equivalent  parts,  by  lines  drawn  from  any  point  in  its 
perimeter  selected  at  random. 

886.  It  is  required  to  divide  a  given  triangle  into  any  num- 
ber of  equivalent  parts,  by  lines  drawn  from  any  point  selected 
at  random  in  the  triangle. 

887.  It  is  required  to  divide  a  given  triangle  into  any  num- 
ber of  equivalent  parts  by  lines  parallel  to  one  side. 

888.  It  is  required  to  divide  a  given  triangle  into  any  num- 
ber of  parts  whose  areas  shall  be  in  a  given  ratio,  by  lines 
drawn  from  one  vertex. 

889.  It  is   required   to   divide   a   given   triangle   into   any 
number  of  parts,  whose  areas  shall  be  in  a  given  ratio,  by  lines 
drawn  from  any  point  selected  at  random  in  the  perimeter. 

890.  It  is  required  to  divide  a  given  triangle  into  any  num- 
ber of  parts,  whose  areas  shall  be  in  a  given  ratio,  by  lines 
drawn  from  any  point  selected  at  random  in  the  triangle. 

891.  It  is  required  to  divide  a  given  triangle  into  any  num- 
ber of  parts,  whose  areas  shall  be  in  a  given  ratio,  by  lines 
drawn  parallel  to  one  side. 

892.  It  is  required  to  divide  a  given  parallelogram  into  any 


164  PLANE  GEOMETRY. 

number  of  equal  parts  by  lines  drawn  parallel  to  one  pair 
of  sides. 

893.  It  is  required  to  divide  a  given  parallelogram  into 
any  number  of  parts,  whose  areas  shall  be  in  a  given  ratio,  by 
lines  parallel  to  one  pair  of  sides. 

894.  It  is  required  to  divide  a  given  parallelogram  into  two 
equivalent  parts  by  a  line  drawn  from  any  point  selected  at 
random  in  the  perimeter. 

895.  It  is  required  to  divide  a  given  parallelogram  into  two 
equivalent  parts  by  a  line  drawn  through  any  point  in  the 
parallelogram  selected  at  random. 

896.  It  is  required  to  divide  a   given   parallelogram   into 
two  parts,  whose  areas  shall  be  in  a  given  ratio,  by  a  line 
drawn  from  one  vertex. 

897.  It  is  required  to  divide  a  given  parallelogram  into  two 
parts,  whose  areas  shall  be  in  a  given  ratio,  by  a  line  drawn 
from  any  point  in  the  perimeter  selected  at  random. 

898.  It  is  required  to  divide  a  given  parallelogram  into  two 
parts,  whose  areas  shall  be  in  a  given  ratio,  by  a  line  drawn 
from  any  point  in  the  parallelogram  selected  at  random. 

899.  It  is   required    to    divide   a  parallelogram   into  two 
equivalent  parts  by  a  line  drawn  parallel  to  a  given  line. 

900.  It  is  required  to  divide  a  parallelogram  into  two  parts, 
whose  areas  shall  be  in  a  given  ratio,  by  a  line  drawn  parallel 
to  a  given  line. 

901.  It  is  required  to  divide  a  parallelogram  into  any  num- 
ber of  equivalent  parts  by  lines  drawn  from  either  vertex. 

902.  It  is  required  to  divide  a  parallelogram  into  any  num- 
ber of  equivalent  parts  by  lines  drawn  from  any  point  in  its 
perimeter  selected  at  random. 

903.  It  is  required  to  divide  a  parallelogram  into  any  num- 


PLANE  GEOMETRY.  165 

her  of  equivalent  parts  by  lines  drawn  from  any  point  in  the 
parallelogram  selected  at  random. 

904.  It  is  required  to  divide  a  parallelogram  into  any  num- 
ber of  equivalent  parts  by  lines  parallel  to  a  given  line. 

905.  It  is  required  to  divide  a  parallelogram  into  any  num- 
ber of  parts,  whose  areas  shall  be  in  a  given  ratio,  by  lines 
parallel  to  a  given  line. 

It  is  required  to  divide  a  trapezoid  into  two  equivalent 
parts  by  a  line  drawn 

906.  Parallel  to  the  bases. 

907.  Perpendicular  to  the  bases. 

908.  Parallel  to  one  of  the  legs. 

909.  Through  one  of  its  vertices. 

910.  Through  a  given  point  in  one  of  its  bases. 

911.  Through  any  point  selected  at  random  in  its  perimeter. 

912.  Through  any  point  selected  at  random  in  the  trapezoid. 

913.  Parallel  to  a  given  line. 

It  is  required  to  divide  a  trapezoid  into  any  number  of 
equivalent  parts  by  lines  drawn 

914.  Parallel  to  the  bases. 

915.  Perpendicular  to  the  bases. 

916.  Parallel  to  one  of  its  legs. 

917.  Through  one  of  its  vertices. 

918.  Through  any  point  selected  at  random  in  one  of  its 
bases. 

919.  Through  any  point  selected  at  random  in  its  perimeter. 

920.  Through  any  point  selected  at  random  in  the  trapezoid. 

921.  Parallel  to  a  given  line. 


166  PLANE  GEOMETEY. 

It  is  required  to  divide  a  given  trapezoid  into  any  number 
of  parts  whose  areas  shall  be  in  a  given  ratio  by  lines  drawn 

922.  Parallel  to  the  bases. 

923.  Perpendicular  to  the  bases. 

924.  Parallel  to  one  of  the  legs. 

925.  Through  either  vertex. 

926.  Through  any  point  selected  at  random  in  one  of  the 
bases. 

927.  Through  any  point  selected  at  random  in  its  perimeter. 

928.  Through  any  point  selected  at  random  in  the  trapezoid. 

929.  Parallel  to  a  given  line. 

930.  It  is  required  to  divide  a  trapezium  into  two  equivalent 
parts  by  a  line  drawn  from  either  vertex. 

931.  It  is  required  to  divide  a  trapezium  into  two  equivalent 
parts  by  a  line  drawn  from  any  point  selected  at  random  in 
its  perimeter. 

932.  Given  the  diameter  of  a  circle,  it  is  required  to  con- 
struct a  straight  line  equal  in  length  to  the  circumference. 

From  Theorem  519  it  is  evident  that  it  can  only  be  approxi- 
mated. 

\Q 

A* 


~"~         ' 


Let  AB  be  the  given  diameter  and  C  its  middle  point. 
Extend  AB  indefinitely  as  AS.  Make  BD  and  DE  each  equal 
to  AB.  At  E  erect  the  perpendicular  EQ,  and  on  it  make 
EFaud  FG  each  equal  to  AB.  Join  AG,  AF,  DG,  and  DF. 
Lay  off  EH  and  HK  each  equal  to  AG,  and  from  K  lay  off 


PLANE  GEOMETRY.  167 

KL  equal  to  AF.  Again,  from  L  make  LM  equal  to  DGr,  and 
MN  equal  DF.  Bisect  EN  v&  P;  bisect  EP  at  72;  and  then 
trisect  -EJjR  at  T  arid  TF.  Then  CT  will  be  the  required  line 
approximately  equal  to  the  circumference  of  the  circle  whose 
diameter  is  AB  ;  for,  calling  the  diameter  unity, 


EL  =  2  CH-  KL  =  2VL3  -  VlO, 


and 

ET  =  TV  (2  Vl3  -  VlO  +  V5  +  V2),  and  therefore 
CT=  2J  +  Jg-  (2V13  -  VlO  +  V5  +  V2)  =  3.1415922  +  . 

Q.E.F. 


168 


PLANE  GEOMETRY. 


APPENDIX. 


1.  The  following  theorem,   and  demonstrations  of  it,  are 
given  as  a  substitute  for  234,  for  those  teachers  who  may 
prefer  it.     Eatio  and  proportion  (261),  however,  should  be 
taken  previous  to  attempting  it. 

2.  In  the  same  or  equal  circles  two  central  angles  are  in  the 
same  ratio  as  the  arcs  which  their  sides  intercept. 


Post.  Let  NBA  and  QKH  be  two  equal  circles,  and  C  and 
I)  two  central  angles. 

We  are  to  prove  Z  C  :  Z  D  : :  Arc  AB  :  Arc  HK 

CASE  I.  —  When  the  angles  are  commensurable. 

Dem.  If  the  angles  are  commensurable,  there  is  some  angle, 
as  W}  which  will  be  contained  an  exact  number  of  times  in 
each. 

Suppose  it  is  contained  m  times  in  Z  (7,  and  n  times  in  Z.  D. 

Then  ZC:ZD::m:n. 

If,  now,  lines  be  drawn  from  C  and  D  dividing  the  two 
angles  into  m  and  n  equal  parts  respectively,  each  part  being 
equal  to  angle  W,  then  arc  AB  will  be  divided  into  m  equal 


PLANE  GEOMETEY.  169 

arcs,  and  HE  into  n  equal  arcs,  the  divisions  all  being  equal, 
by  Theorem  191. 

.*.  Arc  AB  :  Arc  HK :  :  ra  :  n. 

.-.  Z  C  :  Z  D  : :  Arc  AB  :  Arc  HK  (Theorem  288.) 
CASE  II.  —  When  the  angles  are  incommensurable. 


Post.  Let  C  and  D  be  two  incommensurable  central  angles 
in  equal  circles. 

We  are  to  prove    Z  CiZDnArcAB:  Arc  HK 

Conceive  them  to  be  applied  as  in  I. 

Then  if  arc  HKis  not  the  fourth  term  of  this  proportion, 
some  other  arc  greater  or  less  than  HK  must  be.  Suppose 
it  to  be  greater  as  A  W. 

Then  Z  ACB  :  Z  ACK:  :  Arc  AB  :  Arc  A  W.  (a) 

Now  conceive  the  arc  AB  to  be  divided  into  equal  parts  by 
continued  bisection  until  each  part  is  less  than  KW.  Then 
there  must  be  at  least  one  point  of  division  between  K  and 

W,  as  N. 

.-.  Z  ACB  :  Z  ACN:  :  Arc  AB  :  Arc  AN.  (b) 

.'.  Z  ACK :  Z  ACN:  :  Arc  AW:  Arc  AN. 

(Theorem  300.) 

But  the  arc  AN  is  less  than  the  arc  AW.  Consequently, 
if  the  proportion  be  a  true  one,  the  angle  ACN  must  be  less 


170  PLANE  GEOMETRY. 

than  the  angle  ACK.  On.  the  contrary,  it  is  greater,  and 
therefore  the  proportion  cannot  be  a  true  one.  Therefore  the 
supposition  on  which  the  argument  was  based ;  viz.  that  the 
fourth  term  must  be  an  arc  greater  than  HK,  cannot  be  true. 
A  similar  argument  will  prove  that  it  cannot  be  less. 
Consequently  it  must  be  the  arc  HK. 

.-.  ZC:^D::AicAB:  Arc  HK  Q.E.D. 

3.  This  really  means  the  same  thing  as  234 ;  viz.  that  the 
numerical  measure  of  an  angle  at  the  centre  of  a  circle  is 
the  same  as  the  numerical  measure  of  its  intercepted  arc,  if 
the  adopted  unit  of  angle  is  the  angle  at  the  centre  which 
intercepts  the  adopted  unit  of  arc. 

4.  Another  favorite  method  of  demonstrating  this  theorem 
is  that  of  the  method  of  limits,  hereto  appended. 


THEOKY  OP  LIMITS, 

5.  A  constant  quantity,  or  simply  a  constant,  is  a  quantity 
whose  value   remains   unchanged  throughout  the   same   dis- 
cussion. 

6.  A  variable  quantity,  or  simply  a  variable,  is  a  quantity 
which  may  assume  different  values  in  the  same  discussion, 
according  to  the  conditions  imposed. 

7.  The   limit  of  a  variable  is  a  constant  quantity,  which 
the  variable  is  said  to  approach  in  value  whenever  a  regular 
and  definite  increase  or  decrease  in  value  is  assigned  to  the 
latter. 

8.  Whenever  it  can  be  shown  that  the  value  of  a  variable, 
by  such  constant  increase  or  decrease  in  value,  can  be  made  to 
differ  from  that  of  its  limit  by  less  than  any  appreciable  or 
assignable  quantity,  however  small,  this  variable  is  said  to 
approach  indefinitely  to  its  limit. 


PLANE  GEOMETRY.  171 

9.  For  example  : 

A-,  -  1  -  1  -  1  —  KB 
C  D      H  K 

Suppose  a  point  move  from  A  toward  B  under  the  condition 
that  during  the  first  second  it  shall  move  over  one-half  the 
distance  AB,  or  AC,  and  that  during  each  successive  second  it 
shall  move  over  one-half  the  remaining  distance.  Then  at 
the  end  of  the  second  second  it  would  be  at  D,  at  the  end  of 
the  third  at  H,  at  the  end  of  the  fourth  at  K,  and  so  on.  It 
is  evident  that  it  can  never  reach  the  point  B,  for  there  will 
constantly  remain  one-half  the  distance  ;  but  if  its  motion  be 
continued  indefinitely,  it  will  approach  indefinitely  near  to  B. 

Consequently,  the  distance  from  A  to  the  moving  point  is 
an  increasing  variable,  and  AB  is  its  limit;  while  the  distance 
from  B  to  the  moving  point  is  a  decreasing  variable,  with  zero 
as  its  limit. 

10.  Other  illustrations  may  be  given  ;  e.g. 

0.3333  +  ...  =  A  +  Tf¥  +          +    ^n  +.., 


Here  the  sum  of  the  series  of  fractions  is  the  increasing 
variable,  and  approaches  -|  as  its  limit. 

Again  :  let  ABC  be  a  right  triangle,  with 
C  the  right  angle,  and  consider  the  point  B 
to  move  toward  (7;  the  angle  A  will  then 
be  a  decreasing  variable  approaching  zero  as 
its  limit,  and  the  angle  B  will  be  an  increas- 
ing variable  approaching  a  right  angle  as 
its  limit. 


11.    Theorem.     If  two  variables  are  always  equal,  and  each 
approaches  a  limit,  their  limits  are  equal. 

D  Q 


D' 


172  PLANE  GEOMETRY. 

Post.  Let  AB  and  A'B'  be  the  limits  to  which  the  two  equal 
variables  AD  and  A'D'  indefinitely  approach. 

We  are  to  prove  AB  =  A'B'. 

Dem.  AB  and  A'B'  are  either  equal  or  unequal*  Let  us 
suppose  them  unequal,  and  that  AB  is  the  greater.  Mark  off 
AQ  equal  to  A'B'. 

Then  the  variable  A'D'  cannot  exceed  A'B',  but  the  variable 
AD  may  exceed  AQ,  and  consequently  the  variable  AD  be- 
comes greater  than  the  variable  A'D'.  This,  however,  is  con- 
trary to  the  hypothesis  that  the  two  variables  must  always 
be  equal.  Therefore  AB  and  A'B'  cannot  be  unequal ;  i.e. 
they  are  equal.  Q.E.D. 

12.  Theorem.  If  two  variables  are  in  a  constant  ratio,  their 
limits  are  in  the  same  ratio. 


Post.   Let    AD  and   AH  be   two   unequal  variables,   and 
approaching  their  respective  limits  AB  and  AC,  and  having 

AD 

a  constant  ratio  such  that  -  =  m. 


We  are  to  prove  that  =  41*, 


or  that  AD  :  AH  ::AB:AC. 

Dem.   Since      —  —  =±  m,      AD  ==  m  x  AH. 
AH 

And  since  m  X  AH  will  vary  as  AH  varies*  AD  and  m  x  AH 


PLANE  GEOMETRY. 


173 


are   two   equal  variables,  and  therefore,  from   the  previous 
theorem, 

Limit  of  AD  =  Limit  of  m  X  AH,  or 

Limit  of  AD  =  mx  Limit  of  AH. 

Limit  of  AD  _ 
'  Limit  of  AH~ 

but  the  limit  of  AD  is  AB,  and  the  limit  of  AH  is  AC. 
AB 


and  since 


or 


AC 
AD 
AH 
AB 

AC 


—    110   , 

=  w; 
AD 


AH' 
AD:AH::AB:AC. 


Q.E.D. 


13.   Let  us  now  apply  these  principles  to  the  demonstration 
of  the  theorem  enunciated  at  the  beginning  of  the  Appendix. 
In  Case  I.  the  demonstration  will  remain  unchanged. 
Case  II.,  when  the  arcs  are  incommensurable. 


Post.  Let  AQB  and  NHK  be  two  equal  circles,  and  C  and 
D  two  central  angles  whose  arcs  AB  and  HK  are  incom- 
mensurable. 


174  PLANE  GEOMETRY. 


Z  C     Arc  AB 
We  are  to  prove          —  = 


or  Z  C:  Z  D  :  :  Arc  AB  :  Arc 

Ztera.  Conceive  the  arc  AB  to  be  divided  into  any  number 
of  equal  arcs,  and  one  of  these  arcs  to  be  applied  as  a  unit 
of  measure  to  the  arc  HK.  It  will  be  contained  a  certain 
number  of  times  with  a  remainder,  SK,  less  than  the  unit 
of  measure. 

Construct  DS. 

Then,  since  AB  and  HS  are  commensurable, 

Z  C 


Z  D     Arc  HS 

If,  now,  the  number  of  equal  parts  into  which  the  arc  AB 
is  divided  be  indefinitely  increased,  the  unit  of  measure  of  the 
arc  HK  will  be  correspondingly  diminished,  and  the  point 
S  will  get  indefinitely  near  to  K.  Consequently  the  arc 
HS  approaches  indefinitely  to  HK,  and  the  Z  HDS  to  the 
Z.HDK. 

Consequently  the  variables 

its  limit 


and  m  h 


Arc 


Arc  .##  Arc 


But  Arc 


Z  J^Z)^      Arc 
Z  ACB      Arc 


Ai-cHK' 

because  if  two  variables  are  always  equal  and  approaching 
their  limits,  their  limits  are  equal.  Q.E.D. 

14.   Theorems  406  and  313  may  be  Demonstrated  by  similar 
methods. 


PLANE  GEOMETRY. 


175 


SYMMETRY, 

15.  Two  points  are  said  to  be  symmetrical  with  respect  to 
a  point  when  they  are  equidistant  from,  and  in  the  same  line 
with,  this  point.     This  point  is  called  the  centre  of  symmetry. 

16.  Two  points  are  said  to  be  symmetrical  with  respect  to 
a  line  when  the  line  that  joins  them  is  perpendicular  to,  and 
bisected  by,  this  line.     This  line  is  called  an  axis  of  symmetry. 

17.  Two  points  are  said  to  be  symmetrical  with  respect  to 
a  plane  when  the  line  that  joins  them  is  perpendicular  to, 
and  bisected  by,  this  plane.     This  plane  is  called  a  plane  of 
symmetry. 

18.  The  distance  of  either  of  two  symmetrical  points  from 
the  centre  of  symmetry  is  called  the  radius  of  symmetry. 

19.  Two  plane  figures  are  symmetrical  with  respect  to  a 
centre,  axis,  or  a  plane,  when'  any  point  in  either  figure  selected 
at  random  has  a  correspondingly  symmetrical  point   in  the 
other. 

K 


\ 


\ 


JET' 


Fig.  I. 


3 

Fig.  II. 


Fig.  III. 


Fig.  IV. 


176 


PLANE  GEOMETRY. 


C 


•TT 


N 


M 


P' 


Fig.  V. 


Fig.  VI. 


Thus,  in  Figs.  I.,  II.,  and  III.,  the  lines  AB  and  QD  are 
symmetrical,  with  respect  to  the  centre  (7,  the  axis  /£$,  and 
the  plane  MN,  respectively.  In  Figs.  IV.,  V.,  and  VI.,  the 
same  is  true  of  the  triangles  ABQ  and  HR  W. 

20.   A  plane  figure  is  symmetrical 

I.  when  it  can  be  divided  by  an  axis  into  two  figures  sym- 
metrical with  respect  to  that  axis ; 

II.  when  it  has  a  centre  such  that,  if  a  line  be   drawn 
through  it  in  any  direction  at  random,  the  two  points  at  which 
it  intersects  the  perimeter  are  symmetrical  with  respect  to 
that  centre. 


Thus  figure  ABCDH  is  symmetrical  with  respect  to  the 
axis  K/S,  and  ABQDKH  with  respect  to  the  centre  (7.  In 
the  latter  case  PP1  or  NN'  is  called  a  diameter  of  symmetry. 
(See  18.) 


PLANE  GEOMETRY. 


177 


21.   A  geometrical  solid  is  symmetrical 

I.  when  it  can  be  divided  by  a  plane  into  two  solids  sym- 
metrical with  respect  to  that  plane ; 

II.  when  it  has  a  centre  such  that,  if  a  line  be   drawn 
through  it  in  any  direction  selected  at  random,  the  two  points 
at  which  it  intersects  the  surface  are  symmetrical  with  respect 
to  that  centre. 

z 


Fig.  II. 


Fig.  I. 


Thus  figure  ABCDfiFHK,  Fig.  I.,  is  divided  by  the  plane  XZ 
at  the  lines  LM,  MN,  NP,  and  PL,  into  two  figures,  symmet- 
rical with  respect  to  the  plane  XZ.  Hence  it  is  symmetrical. 

Similarly,  in  Fig.  II.,  the  points  P  and  P  being  symmetrical 
with  respect  to  the  point  (7,  according  to  above  definition,  the 
figure  is  symmetrical. 

THEOREMS. 

22.  The  centre  of  a  circle  is  a  centre  of  symmetry. 

23.  The  diameter  of  a  circle  is  an  axis  of  symmetry. 

24.  The  line  which  bisects  the  vertical  angle  of  an  isosceles 
triangle  is  an  axis  of  symmetry. 

25.  Either  altitude  of  an  equilateral  triangle  is  an  axis  of 
symmetry. 

26.  The  point  of  intersection  of  two  altitudes  of  an  equi- 
lateral triangle  is  a  centre  of  symmetry. 


178  PLANE  GEOMETRY. 

27.  A  segment  of  a  circle  is  a  symmetrical  figure. 

28.  That  part  of  a  circle  included   between  two   parallel 
chords  is  a  symmetrical  figure. 

29.  The  common  part  of  two  intersecting  circles  is  a  sym- 
metrical figure. 

30.  The  diagonal  of  a  square  is  an  axis  of  symmetry. 

31.  Every  equilateral  tetragon  is  a  symmetrical  figure. 
(How  many  axes  of  symmetry  does  a  square  have  ?) 

32.  The  point  of  intersection  of  the  diagonals  of  a  parallelo- 
gram is  a  centre  of  symmetry. 

33.  An  isosceles  trapezoid  is  a  symmetrical  figure. 

34.  If  one  diagonal  of  a  tetragon  divides  it  into  two  isosceles 
triangles,  the  other  diagonal  is  an  axis  of  symmetry. 

35.  The  bisector  of  an  angle  of  a  regular  polygon  is  an  axis 
of  symmetry. 

36.  The  perpendicular  bisector  of  one  side  of  a  regular  poly- 
gon is  an  axis  of  symmetry. 

37.  If  the   angles   at   the   extremities  of  one   side  of  an 
equilateral  pentagon  be  equal,  the  pentagon  is  a  symmetrical 
figure. 

38.  If  two  diametrically  opposite  angles  of  an  equilateral 
hexagon  are  equal,  the  hexagon  is  a  symmetrical  figure. 

39.  Every  equiangular  tetragon  is  a  symmetrical  figure. 

40.  If  a  figure  have  two  axes  of  symmetry  perpendicular 
to  each  other,  their  intersection  is  a  centre  of  symmetry. 

Post.  Let  ABDH,  etc.,  be  a  figure  having  the  two  axes  of 
symmetry  PP'  and  MM'  _L  to  each  other. 

We  are  to  prove  that  their  point  of  intersection  C  is  a 
centre  of  symmetry. 


PLANE  GEOMETRY. 


179 


Cons.  From  any  point  in  the  perimeter  selected  at  random, 
as  Q,  construct  QOA.MM',  QF±PP',  and  join  TL,  OC,  and 
FC. 

Dem.  OT=TQ  =  LC.  Why  ? 


,TT 


Then  what  kind  of  a  figure  is  OTLC?  What  relation,  then, 
between  TL  and  OC  ?  Compare  in  a  similar  manner  TL  and 
OF.  Finally  compare  OC  and  CF. 

Hence  points  0  and  F  are  in  the  same  line  with  and  equi- 
distant from  C.  Hence  C  is  a  centre  of  symmetry.  Q.E.D. 


YB  1 7298 


